Spatial Reasoning · Additional Mathematics

1Introductionp. 62

A parabola is a curve with applications in fields such as physics, engineering, and economics. Whether it is the path of a projectile, the shape of a satellite dish, or the beam from a flashlight, parabolas effectively model both natural and man-made phenomena. In this chapter, you will learn about the features of parabolas in detail. You will learn how to sketch and analyse parabolas derived from quadratic functions and geometric definitions. This section will equip you to:

Construct a parabola from a given quadratic equation of the form $y = ax^2 + bx + c$ and identify and explain its key features, such as the vertex, axis of symmetry, focus, directrix, and direction of opening.
Sketch a parabola, and from its symmetry and shape, derive the general geometric relationship that links a parabola's focus, directrix, and locus definition.
Sketch a parabola given the directrix and focus, using the formal geometric definition: a parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). You will learn how to plot this step-by-step and derive its corresponding equation.
Derive the equations of the tangent and normal to a parabola at a given point on the curve.
Apply differentiation and geometric principles to interpret these lines and understand their relationship with the parabola.

KEY IDEAS

• Definition of a Parabola: A parabola is the set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix). • Key Features of a Parabola o Vertex: The highest or lowest point of the parabola. o Axis of Symmetry: A vertical or horizontal line that passes through the vertex and divides the parabola into two mirror-image halves. o Focus: A point inside the parabola that defines its shape. o Directrix: A fixed line outside the parabola used in its geometric definition. o Focal Length: The distance from the vertex to the focus (or to the directrix).

• Forms of the Parabola o Vertical (opens up or down): $(x - h)^2 = 4p(y - k)$ o Horizontal (opens left or right): $(y - k)^2 = 4p(x - h)$ • Sketching a Parabola: A parabola can be sketched by o Identifying the vertex, axis of symmetry, and direction of opening. o Plotting key points (including the focus and directrix if available). o Drawing a smooth, symmetric curve. • Parabola from Focus and Directrix o Given a focus and directrix, the parabola can be constructed by plotting all points equidistant from them. o The equation can then be derived using the distance formula. • Tangents and Normals o The tangent to a parabola at a point touches it at exactly one point and has the same slope as the curve at that point. o The normal is perpendicular to the tangent and also passes through that point. o Both equations can be deduced using calculus or coordinate geometry methods.

2Describing the Shape of the Graph of Quadratic Functionsp. 63

The general quadratic function is of the form $f(x) = y = ax^2 + bx + c$ , where $a, b,$ and $c$ are constants and $a \neq 0.$ Also, $x$ and $y$ are variables. The basic form of the quadratic function is $y = x^2$ . [Figure] Figure 3.1: Graph of $y = x^2$

Another name for the quadratic function is parabola. A parabola is an example of a conic section. Other examples of conic sections are circles, ellipses, and hyperbolas. These shapes are called conics because they are generated from cones. A parabola is generated when a plane intersects a cone parallel to the generating line (slant height) as illustrated below. [Figure] Figure 3.2: Parabola generating process

Activity 3.1 Discussing Real-life examples of Parabolas

Examine each image of real-life parabolas provided. a. Projectile Motion (Throwing or Launching Objects) The path of a ball when kicked or thrown follows a parabola. [Figure] Figure 3.3: Motion of a projected ball b. Satellite Dishes Satellite dishes are shaped like paraboloids (3D parabolas) so they can focus signals to a single point — the receiver. This design ensures maximum signal strength. [Figure] Figure 3.4: Satellite dish

c. Car Headlights and Flashlights The reflector inside is shaped like a parabola so that light rays from the bulb at the focus are reflected in parallel rays. This creates a strong, directed beam. [Figure] Figure 3.5: An automobile headlight d. Bridges and Arches Some suspension bridge cables and arch supports are shaped like parabolas [Figure] Figure 3.6: A bridge e. Water Fountains Water shooting from a fountain often follows a parabolic trajectory due to gravity. The shape depends on the angle and force of the water jet. [Figure] Figure 3.7: Water fountain 2. Note the vertex position, direction of opening (upward or downward), and any visible symmetry. 3. Record observations in your notebook. 4. Share observations about the images with a peer or your teacher 5. Decide which image represents the clearest example of a parabola and explain why. 6. Discuss with a peer how the vertex and orientation of the parabola influence the object’s function. 7. Select one image that interests you the most.

Sketch a possible quadratic function that represents it on graph paper or using a digital tool.
Label the vertex, axis of symmetry, and orientation.
Write a brief description (3–5 sentences) of your sketch, explaining how it relates to the real-life object.
Present your sketch and description to the class or group members.
Listen to others’ presentations and ask at least one question or give one positive observation.
Reflect on how understanding parabolas can help in explaining or designing real-life objects.

Shape of a Quadratic Function

The graph of a parabolic function of the form $f(x) = y = ax^2 + bx + c$ is either of the diagrams below. These types of graphs are symmetrical, with the line of symmetry occurring at $x = -\frac{b}{2a}$ [Figure] or [Figure] Figure 3.8: Graph of $y = -x^2$ and $y = x^2$ The orientation of the graph of is dependent on the value of $a$ . If the graph is written in the form $f(x) = a(x - h)^2 + k$ the minimum or maximum value is $k$ and the line of symmetry is $x = h$ or $x - h = 0$ . Also, the vertex of the quadratic graph will be $(h, k)$ .

Maximum graph

If The The $a < 0$ line graph , of the is symmetry increasing graph is is shaped when $x = -\frac{b}{2a}$ $f\left(-\frac{b}{2a}\right)$ like $x < -\frac{b}{2a}$ the intersection symbol [Figure] The graph will have a maximum turning point. The coordinates of the turning point are $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$ and the maximum value is

Figure 3.9: A maximum graph The graph is decreasing when $x > -\frac{b}{2a}$

The graph is increasing when $x < -\frac{b}{2a}$ or $\left(-\infty < x < -\frac{b}{2a}\right)$ The graph is decreasing when $x > -\frac{b}{2a}$ or $\left(-\frac{b}{2a} < x < \infty\right)$ The domain of the graph is $(-\infty < x < \infty) = x \in R$ The range is $y > f\left(-\frac{b}{2a}\right)$ or $\left(-\infty < y < f\left(-\frac{b}{2a}\right)\right)$

Minimum graph

The $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$ If $a > 0$ coordinates , the graph of and the is shaped the turning minimum like point the is value Union is symbol. $f\left(-\frac{b}{2a}\right)$ [Figure] The graph will have a minimum turning point.

The line of symmetry is $x = -\frac{b}{2a}$ Figure 3.10: A minimum graph The graph is increasing when $x > -\frac{b}{2a}$ or $\left(-\frac{b}{2a} < x < \infty\right)$ The graph is decreasing when $x < -\frac{b}{2a}$ or $\left(-\infty < x < -\frac{b}{2a}\right)$ The domain of the graph is $(-\infty < x < \infty) = x \in R$ The range is $y > f\left(-\frac{b}{2a}\right)$ or $\left(f\left(-\frac{b}{2a}\right) < y < \infty\right)$

Example 3.1

Figure 3.11 shows the graph of a quadratic function.

[Figure] Figure 3.11: A quadratic function

a. Describe the shape of the quadratic function. b. Write an equation to represent the function.

Solution

a. The graph opens upwards since it is a minimum graph. The equation of the line of symmetry is $x = 1$ or $x - 1 = 0$ , the minimum point is (1, -4) and the minimum value is -4. The graph is increasing when $1 < x < \infty$ The graph is decreasing when $-\infty < x < 1$ The domain of the graph is $-\infty < x < \infty$ or $x \in R$ The range is $-4 < y < \infty$ b. Using $y = a(x - h)^2 + k$ where (h, k) is the turning point and a is $a$ parameter. From the graph the turning point is $(1, -4)$ . This means, $h = 1$ and $k = -4$ $y = a(x - 1)^2 - 4$ To find $a$ , substitute any other ordered pair that satisfies the equation and solve for $a$ . The curve passes through the point (3, 0) $0 = a(3 - 1)^2 - 4$ $0 = a(2)^2 - 4$ $4 = 4a$ $a = \frac{4}{4} = 1$ Therefore, $y = 1(x - 1)^2 - 4$ Now, expand and simplify $y = (x^2 - 2x + 1) - 4$ $y = x^2 - 2x - 3$

Example 3.2

The path of a projectile is represented by the graph below, where H denotes the height (in metres) at time $t$ seconds, and $t$ is the time elapsed since the projectile was launched.

[Figure] Figure 3.12: Path of a projectile

a. Write a quadratic equation to represent the path of the projectile. b. Describe the shape of the H in F igure 3.12 .

Solution

a. Using $H = a(t - h)^2 + k$ where $(h, k$ ) is the turning point and $a$ is a parameter. From the graph, the turning point is (2.5, 100). This means, $h = 2.5$ and $k = 100$ $H = a(t - 2.5)^2 + 100$ To find $a$ , substitute any other ordered pair that satisfies the equation, and solve for $a$ . The curve passes through the origin, (0, 0) $0 = a(0 - 2.5)^2 + 100$ $0 = a\left(\frac{5}{2}\right)^2 + 100$ $-100 = \frac{25}{4}a$ $a = -100 \times \frac{4}{25} = -16$ Therefore, $H = -16(t - 2.5)^2 + 100$ Now, expand and simplify $H = -16(t^2 - 5t + 6.25) + 100$ $H = -16t^2 + 80t - 100 + 100$

$H = -16t^2 + 80t$ $H = 80t - 16t^2$ b. The graph opens downwards since it is a minimum graph. The equation of the line of symmetry is $t = 2.5$ or $2t - 5 = 0$ . The maximum point is (2.5, 100), and the maximum value is 100. The graph is increasing when $0 < t < 2.5\mathrm{s}$ The graph is decreasing when $2.5\mathrm{s} < t < 5.0\mathrm{s}$ The domain of the graph is $0 < t < 5$ The range is $0 < H < 100$

3Exploring the Graphs of Quadratic Functions with Different Parametersp. 70

Before we start exploring the graphs of quadratic functions, let's simplify the general quadratic equation by completing the square. The general form of a quadratic function is: n is: $f(x) = ax^2 + bx + c$ . By completing the square, we can write it in the form $f(x) = a(x + h)^2 + k$ . The parameters $a, h$ and $k$ determines the shape of the graph of a quadratic function. Let us now explore how the values of $a, h$ and $k$ affect the shape of the graph. Graph of the quadratic function when $\boldsymbol{a = 1}$ , $\boldsymbol{h = 0}$ and $\boldsymbol{k = 0}$ Substituting $a = 1$ , $h = 0$ and $k = 0$ into $f(x) = a(x + h)^2 + k$ , we get $f(x) = 1(x + 0)^2 + 0$ $f(x) = x^2$ , which is the basic quadratic function This means that the basic equation will have a vertex (0, 0) and a line of symmetry $x = 0$ [Figure] Figure 3.13: Graph of $f(x) = x^2$ The graph of $f(x) = x^2$ has the following properties:

It is U-shaped.
The turning point is located at the origin.
The turning point is a minimum point. $\boldsymbol{|a| > 1}$

Activity 3.2 Graph of the quadratic equation when IaI>1, h = 0 and k = 0

Step 1: Substitute $a = 2$ , $h = 0$ and $k = 0$ into $f(x) = a(x + h)^2 + k$ . Your answer should be $f(x) = 2(x + 0)^2 + 0$ $f(x) = 2x^2$ Step 2: Draw $f(x) = x^2$ and $g(x) = 2x^2$ on the same graph. Your answer should be similar to [Figure] Figure 3.14: Graphs of $f(x) = x^2$ and $g(x) = 2x^2$ Step 3: Compare the graph of $f(x) = x^2$ to $g(x) = 2x^2$ and state any observations you make. Your observations should include the following.

They both have the same U-shape.
Their turning points are both located at the origin.
Their turning points are both minimum points.
The graph of $f(x) = x^2$ has been ‘ shrunk’ to obtain the graph of g $(x) = 2x^2$ . This is known as a vertical stretch. Step 4: From your observations in step 3, what conclusion can you make about the parameter $\boldsymbol{a}$ in a quadratic function? Is your conclusion similar to the one below? As the absolute value of $\text{"}a\text{"}$ increases, the graph becomes narrower compared to the basic graph. If your answer is no, recheck your graphs.

Activity 3.3 The graph of the quadratic equation when -1 < a < 1, h = 0 and

k = 0

Step 1: Substitute $a = 0.5$ , $h = 0$ and $k = 0$ into $f(x) = a(x + h)^2 + k$ . Your answer should be $f(x) = 0.5(x + 0)^2 + 0$ $f(x) = 0.5x^2$ Step 2: Draw $f(x) = x^2$ and $g(x) = 0.5x^2$ on the same graph. Your answer should be similar to [Figure] Figure 3.15: Graphs of $f(x) = x^2$ and $g(x) = 0.5x^2$ Step 3: Compare the graph of $f(x) = x^2$ to $g(x) = 0.5x^2$ and state any observations you make. Did you observe the following?

They are all U-shaped.
Their turning points are all located at the origin.
Their turning points are all minimum points.
The graph of $f(x)$ has been stretched vertically (dilated) to obtain graphs of g $(x).$
The graph of $f(x)$ is narrower compared to $g(x)$ If your answer is no, recheck your graphs.

Step 4: From your observations in step 3, what conclusion can you make about the parameter ‘ $\boldsymbol{a}$ ’ in a quadratic function. Is your conclusion similar to the one below? As the absolute value of ‘ $\boldsymbol{a}$ ’ is less than 1, the graph becomes wider as compared to the basic graph. If your answer is no, recheck your graphs.

What happens when the values of ‘ $\boldsymbol{a}$ ’ is negative in all the situations that have been discussed? Let us graph them and do comparisons. [Figure] Figure 3.16: Graphs of $f(x)$ , $g(x)$ and $h(x)$ and their reflections It can be observed from Figure 3.16 that when the value of ‘ $a$ ’ is negative, the resulting graph is a reflection of the original graph about the $x -$ axis. The following conclusions can be made from the graphs of Figure 3.16:

The greater the value of $|a|$ , the narrower the curve.
The smaller the value of $|a|$ , the wider the curve.
If $a$ is negative, the resulting graph is a reflection about the $x -$ axis.

Activity 3.4 Graph of the quadratic function when h is any number other

than 0

a. When $\boldsymbol{h}$ is a positive number Step 1: Substitute $h = 2$ , $a = 1$ and $k = 0$ into $f(x) = a(x + h)^2 + k$ Was your answer similar to the one below? $f(x) = (x + 2)^2$

Step 2: Draw $f(x) = x^2$ and $g(x) = (x + 2)^2$ on the same graph

[Figure] Figure 3.17: Graphs of $f(x) = x^2$ and $g(x) = (x + 2)^2$ Step 3: Compare the graph of $f(x) = x^2$ to $g(x) = (x + 2)^2$ and state any observations you make. Did you observe the following? They are both U-shaped. While the turning point of $f(x)$ is located at the origin, the turning point of $g(x)$ is located at $(-2, 0)$ indicating a horizontal shift of the graph of $f(x)$ two units to the left.

Their turning points are both minimum points. The graphs of $f(x)$ and $g(x)$ do not change in size. There is no shrinkage or dilation (i.e., no stretching); they are merely translated. If no, recheck your graphs

Step 4: From your observations in step 3, what conclusion can you make about the parameter $\boldsymbol{h}$ in a quadratic function. Is your conclusion similar to the one below? When h is positive, the graph shifts h units to the left or translated by the vector $\begin{pmatrix}-h \\ 0\end{pmatrix}$ . If no, recheck your graphs

b. When $\boldsymbol{h}$ is a negative number Step 1: Substitute $h = -1$ , $a = 1$ and $k = 0$ into $f(x) = a(x + h)^2 + k$ Was your answer similar to the one below? $f(x) = (x - 1)^2$

Step 2: Draw $f(x) = x^2$ and $g(x) = (x - 1)^2$ on the same graph [Figure] Figure 3.18: Graphs of $f(x) = x^2$ and $g(x) = (x + 1)^2$ Step 3: Compare the graph of $f(x) = x^2$ to $g(x) = (x - 1)^2$ and state any observations you make. Did you observe the following? They are both U-shaped. While the turning point of $f(x)$ is located at the origin, the turning point of $g(x)$ is located at $(1, 0)$ indicating a horizontal shift of the graph of $f(x)$ one unit to the right.

Their turning points are both minimum points. The graphs of $f(x)$ and $g(x)$ do not change in size. There is no shrinkage or dilation (i.e., no stretching); they are merely translated. If no, recheck your graphs

Step 4: From your observations in step 3, what conclusion can you make about the parameter $\boldsymbol{h}$ in a quadratic function? Your conclusion should be similar to the one below: When h is negative, the graph shifts h units to the right or is translated by the vector $\begin{pmatrix}h \\ 0\end{pmatrix}$ .

General observation about the value of h

The following conclusions can be made about the effect of changing the value of $h$ in a function defined as $f(x) = a(x + h)^2 + k$ from the graphs in Figure 3.17 through to Figure 3.18:

The value of “h” indicates the distance of horizontal shift of the resulting graph form the graph of $f(x) = x^2$ .
The greater the value of “ $|h|$ ”, the farther the resulting curve is to the graph of $f(x) = x^2$ .
The smaller the value of “ $|h|$ ”, the closer the resulting curve is to the graph of $f(x) = x^2$ .
When $h < 0$ , there is a shift to the right.
When $h > 0$ , there is a shift to the left.

Activity 3.5 Shape of parabolic graphs when the value of “ $\boldsymbol{k}$ ” is not 0

When $\boldsymbol{k}$ is a positive number Step 1: Substitute $k = 4$ , $a = 1$ and $h = 0$ into $f(x) = a(x + h)^2 + k$ Your answer should be same as: $f(x) = x^2 + 4$

Step 2: Draw $f(x) = x^2$ and $g(x) = x^2 + 4$ on the same graph [Figure] Figure 3.19: Graphs of $f(x) = x^2$ and $g(x) = x^2 + 4$ Step 3: Compare the graph of $f(x) = x^2$ to $g(x) = x^2 + 4$ and state any observations you make. Did you observe the following?

They are all U-shaped.
While the turning point of $f(x)$ is located at the origin, the turning point of $g(x)$ is located at $(0, 4)$ indicating a vertical translation of the graph of $f(x)$ 4 units upward
Their turning points are both minimum.
The graphs of $f(x)$ and $g(x)$ do not change in size. There is no shrinkage or dilation; they are simply translations. If no, recheck your graphs.

Step 4: From your observations in step 3, what conclusion can you make about the parameter $\boldsymbol{k}$ in a quadratic function? Did you come to the conclusion below? When $\boldsymbol{k}$ is positive, the graph shifts $\boldsymbol{k}$ units upwards or translated by the vector $\begin{pmatrix}0 \\ k\end{pmatrix}$ . If no, recheck your graphs. When $\boldsymbol{k}$ is a negative number Step 1: Substitute $k = -2$ , $a = 1$ and $h = 0$ into $f(x) = a(x + h)^2 + k$ Your answer should be same as: $f(x) = x^2 - 2$

Step 2: Draw $f(x) = x^2$ and $g(x) = x^2 - 2$ on the same graph [Figure] Figure 3.20: Graphs of $f(x) = x^2$ and $g(x) = x^2 - 2$ Step 3: Compare the graph of $f(x) = x^2$ to $g(x) = x^2 - 2$ and state any observations you make.

Did you observe the following? They are all U-shaped. While the turning point of $f(x)$ is located at the origin, the turning point of $g(x)$ is located at $(0, -2)$ indicating a vertical translation of the graph of $f(x)$ , 2 units downwards. Their turning points are both minimum. The graphs of $f(x)$ and $g(x)$ do not change in size. There is no shrinkage or dilation; they are simply translations. If no, recheck your graphs. Step 4: From your observations in Step 3, what conclusion can you make about the parameter k in a quadratic function? Did you come to the conclusion below? When $k$ is negative, the graph shifts $k$ units downwards or is translated by the vector $\begin{pmatrix}0 \\ -k\end{pmatrix}$ . If no, recheck your graphs.

General observation about the value of k

The following conclusions can be made about the effect of changing the value of “ $k$ ” in a function defined as $f(x) = a(x + h)^2 + k$ from the graphs in Figure 3.19 and Figure 3.20:

The value of “ $|k|$ ” indicates the distance of vertical shift of the resulting graph form the graph of $f(x) = x^2$ .
The greater the value of “ $|k|$ ”, the farther the resulting curve is to the graph of $f(x) = x^2$ .
The smaller the value of “ $|k|$ ”, the closer the resulting curve is to the graph of $f(x) = x^2$ .
When $k < 0$ , there is a downward shift.
When $k > 0$ , there is an upward shift.

Example 3.3

Figure 3.21 shows a graph of a quadratic function. [Figure] Figure 3.21: Graph of a quadratic function

a. Write the quadratic function in the form $y = a(x - h)^2 + k$ where a, h, and k are parameters. b. Describe the shape of the graph when compared to $y = x^2$

Solution

a. The turning point is (-1, 4). This implies that $h = -1$ and $k = 4$ $y = a(x + 1)^2 + 4$ Since (1, 0) satisfies the graph, substitute it into the equation and solve for $a$ $0 = a(1 + 1)^2 + 4$ $0 = 4a + 4$ $-4 = 4a$ $a = -\frac{4}{4} = -1$ This means that $y = -(x + 1)^2 + 4$ . b. The graph of $y = -(x + 1)^2 + 4$ results from reflecting the graph of $y = x^2$ in the y-axis, followed by shifting it one unit to the left and 4 units upwards, or a translation by the vector $\begin{pmatrix}-1 \\ 4\end{pmatrix}.$ [Figure] Figure 3.22: Graphs of $y = -(x + 1)^2 + 4$ and $y = x^2$

Example 3.4

The diagram below shows a graph of a quadratic function. [Figure] Figure 3.23: Graphs of quadratic function

a. Write the quadratic function in the form $y = a(x - h)^2 + k$ where a, h and k are parameters. b. Describe the shape of the graph as compared to $y = x^2$

Solution

The turning point is (2, 0.5). This implies that $h = 2$ and $k = 0.5$ $y = a(x - 2)^2 + 0.5$ Since (0, 1.5) satisfies the graph, substitute it into the equation and solve for $a$ $1.5 = a(0 - 2)^2 + 0.5$ $1.5 - 0.5 = 4a$ $1 = 4a$ $a = \frac{1}{4} = 0.25$ This means that $y = \frac{1}{4}(x - 2)^2 + 0.5$ a. The graph of $y = \frac{1}{4}(x - 2)^2 + 0.5$ results from shifting the graph of $y = x^2$ two units to the right and 0.5 units upwards, or a translation by the vector $\begin{pmatrix}2 \\ 0.5\end{pmatrix}$ . Also, to obtain the graph of $y = \frac{1}{4}(x - 2)^2 + 0.5$ , the graph of $y = x^2$ is widened. [Figure] Figure 3.24: Graphs of $y = \frac{1}{4}(x - 2)^2 + 0.5$ and $y = x^2$

4Sketching Quadratic Graphsp. 81

Step 1: Express the function in the form $f(x) = a(x - h)^2 + k.$ When expressed in this form, (h, k) is the turning point or vertex. If a>0, the curve has a minimum and opens upwards. If a<0, the curve has a maximum and opens downwards. Step 2: Sketch the function by comparing it to $f(x) = x^2$ Alternatively, Step 1: Determine the shape. Either it is opening upwards, or opening downwards. Step 2: Find the vertex or the turning point. This is given as $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$ Step 3: Find the $x$ -intercept(s), if any. For $x$ -intercept put $f(x) = y = 0$ . Step 4: Find the $y$ -intercept. To find the $y$ -intercept, put $x = 0$ Step 5: Connect these points to form the curve.

Example 3.5

Sketch the graph $y = x^2 + 6x + 5$ and compare your graph to $y = x^2$ . Solution Step 1: Write $y = x^2 + 6x + 5$ in the form $y = a(x - h)^2 + k$ $y = x^2 + 6x + 5$ $y = x^2 + 3x + 3x + 9 - 9 + 5$ $y = x(x + 3) + 3(x + 3) - 4$ $y = (x + 3)^2 - 4$ Comparing to $y = a(x - h)^2 + k$ , we have $a = 1, h = -3$ and $k = -4$ This means that the turning point is $(-3, -4).$ Since a is positive, the graph opens upwards (minimum graph). This is a translation of the graph of $y = x^2$ by moving 3 units to the left and 4 units down, or the vector $\begin{pmatrix}-3 \\ -4\end{pmatrix}$ . [Figure] Figure 3.25: Graphs of $y = (x + 3)^2 - 4$ and $y = x^2$

Example 3.6

Sketch the graph $y = 7 - 4x - 2x^2$ and compare your graph to $y = x^2$ .

Solution

First, write $y = 7 - 4x - 2x^2$ in the form $y = a(x - h)^2 + k$ $y = -2x^2 - 4x + 7$ (factor $-2$ by dividing each term by $-2$ ) $= -2(x^2 + 2x - 3.5)$ (complete the square) $= -2(x^2 + x + x + 1 - 1 - 3.5)$ $= -2[(x + 1)^2 - 4.5]$ $= -2(x + 1)^2 + 9$ Comparing $y = -2(x + 1)^2 + 9$ to $y = a(x + h)^2 + k$ shows that $a = -2, h = -1$ and $k = 9$ . This means that (-1, 9) is the turning point. Since $a$ is negative, the graph is maximum. So, the graph will open downwards. The graph of $y = -2(x + 1)^2 + 9$ is a reflection of $y = x^2$ in the x-axis, followed by a shift of 1 unit to the left and 9 units up, or a translation by the vector $v = \begin{pmatrix}-1 \\ 9\end{pmatrix}$ . Also, since |a|=2, the graph of $y = -2(x + 1)^2 + 9$ will be a vertical shrink of $y = x^2$ . [Figure] Figure 3.26: Graphs of $y = x^2$ and $y = -2x^2 - 4x + 7$

Example 3.7

Describe and sketch the graph of $y = -2x^2 + 4$ .

Solution

$y = -2x^2 + 4$ is the same as $y = -2(x - 0)^2 + 4$ Compared to $y = a(x + h)^2 + k$ shows that $a = -2, h = 0$ and $k = 4$

The vertex is at (0, 4) and the line of symmetry is x=0 To obtain the graph of $y = -2x^2 + 4$ , the graph of $y = x^2$ is shrunk (vertically stretched), reflected about the $x -$ axis and shifted upward by four units. [Figure] Figure 3.27: Graph of $y = -2x^2 + 4$ and $y = x^2$

5Describing a Given Locus as a Parabolap. 83

Earlier, you learned that the graph of the quadratic function $f(x) = ax^2 + bx + c$ is a parabola that opens upward or downward. The following definition of a parabola is more general in the sense that it is independent of the orientation of the parabola.

Definition of Parabola

A parabola is the set of all points $(x, y)$ in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. [Figure] Figure 3.27: A parabola as a locus of points In Figure 3.27 , the red curve illustrates a parabola. Points $\mathrm{V, A,}$ and $\mathrm{B}$ lie on the curve, where $\mathrm{V}$ represents the vertex or turning point of the parabola. The point labelled $\mathrm{F}$ is the focus, and the dotted horizontal line indicates the directrix.

By definition, a parabola is the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). This means that, for any point chosen on the parabola, the distance to the focus is exactly equal to its perpendicular distance to the directrix. So, $|AF| =$ distance between $\mathrm{A}$ and the directrix. Likewise, $|\mathrm{BF}| =$ perpendicular distance between $\mathrm{B}$ and the directrix. Also, $|FV| =$ distance between $\mathrm{V}$ and the directrix.

The Focus

The focus is a fixed point inside the parabola. Every point on the parabola is equidistant from the focus and a fixed line called the directrix. The focus lies on the axis of symmetry. It helps determine the shape and direction of the parabola. The vertex is always midway between the focus and the directrix.

Directrix of a Parabola

The directrix of a parabola is a fixed straight line used in the definition of the parabola. Every point on the parabola is equidistant from a fixed point called the focus and the directrix. It lies opposite the focus, on the other side of the vertex. The vertex of the parabola is exactly halfway between the focus and the directrix. It helps determine the shape and orientation of the parabola.

Line of symmetry of a Parabola

Also called the axis of symmetry, it is a vertical or horizontal line that divides the parabola into two mirror-image halves and passes through the vertex and the focus.

Focal Length of a Parabola (p)

This is the distance between the vertex and the focus (or the vertex and the directrix, since they are equidistant). It tells us how "wide" or "narrow" the parabola is. The larger the focal length, the wider the parabola.

Vertex

This is the turning point or the point where the parabola changes direction. It is located halfway between the focus and the directrix and lies on the axis of symmetry.

Opening direction

A parabola can open in different directions depending on its equation A parabola of the form $y = ax^2$ A parabola of the form $x = ay^2$ Opens upward if $a > 0$ Opens right if $a > 0$ Opens downward if $a < 0$ Opens left if $a < 0$

6Describing the Features of a Parabolap. 85

A parabola is a U-shaped curve that can open up, down, left, or right. It has specific key features that help describe its shape, position, and orientation. These key features are the focus, line of symmetry, focal length, vertex, and opening direction. We can also use these key features to sketch the parabola. Let us use the activity below to generate the parabolic equation with vertex (0,0), focal length $\boldsymbol{p}$ , and a directrix which is parallel to the $x\text{-axis}$ .

Activity 3.6 Generating the parabolic equation with vertex (0, 0), focal

length p, and a directrix which is parallel to the x-axis.

Step 1: Make a copy of the graph below The vertex is at the origin (0, 0). The focal length is $p$ , the coordinates of the focus will be $F(0, p)$ and the equation of the directrix will intersect the y-axis at $-p$ , R $(x, y)$ is any point on the parabola. So, a perpendicular from R will intersect the directrix at A $(x, -p)$ [Figure] Figure 3.28: A parabola with vertex $(0, 0)$ and focus $(0, p)$ Step 2: Find the distance between points $R$ and $A$ Your answer should be same as: $A(x, -p)$ and $R(x, y)$ So, $|RA| = \sqrt{\left(y - (-p)\right)^2 + (x - x)^2}$ $= \sqrt{(y + p)^2}$

Step 3: Find the distance between points $R$ and $F$ Your answer should be same as: $F(0, p)$ and $R(x, y)$ So, $|RF| = \sqrt{(y - p)^2 + (x - 0)^2}$ $= \sqrt{(y - p)^2 + x^2}$

Step 4: Equate |RF| to |RA| and simplify the resulting equation.

Your answer should be similar to: $\sqrt{(y - p)^2 + x^2} = \sqrt{(y + p)^2}$ Take square of both sides $(y - p)^2 + x^2 = (y + p)^2$ Let $x^2$ stand alone. $x^2 = (y + p)^2 - (y - p)^2$ Expand RHS using difference of two squares $x^2 = (y + p - y + p)(y + p + y - p)$ Simplify RHS $x^2 = (2p)(2y)$ $x^2 = 4py$ or $4py = x^2$

This gives the basic equation of the parabola where the directrix is parallel to the x-axis.

Example 3.8

Describe the features of the parabola $y = x^2$ Solution $y = x^2$ making we have is y $y = \frac{1}{4p}x^2$ the the subject, basic quadratic ……………. equation (2) [Figure] $y = x^2$ …………….. (1) $4py = x^2$

Comparing equation 1 to equation 2, we Figure 3.29: Features of parabola $y = x^2$ have $1 = \frac{1}{4p}$ $4p = 1$ $p = \frac{1}{4} = 0.25$ This means that the focal length of this parabola is 0.25, The directrix is $y = -0.25$ The focus is (0, 0.25) Figure 3.29 shows the graph of $y = x^2$

Example 3.9

Describe the features of the parabola $y = -x^2$ Solution $y = -\frac{1}{4p}x^2$ making Comparing $y$ the equation subject, ……………. 1 we to equation have (2) 2, we [Figure] $y = -x^2$ …………….. (1) $4py = x^2$

have $1 = -\frac{1}{4p}$ $4p = -1$ Figure 3.30: Features of parabola $y = -x^2$ $p = -\frac{1}{4} = -0.25$ This means that the focal length of this parabola is 0.25, The directrix is $y = 0.25$ The focus is $(0, -0.25)$ Let us use the activity below to generate the parabolic equation with vertex (0, 0), focal length $\boldsymbol{p}$ , and a directrix which is parallel to the y-axis.

Activity 3.7 Generating the parabolic equation with vertex (0, 0), focal

length P, and a directrix which is parallel to the y-axis.

Step 1: Make a copy of the graph below $-p, R(x, y)$ $(-p, x$ The directrix So, from a vertex perpendicular $R$ ) will will is intersect is at intersect any the point origin the the on directrix (0,0). x-axis the parabola. at at A [Figure] The focal length is p, the coordinates of the focus will be F $(p, 0)$ and the equation of the

Figure 3.31: Graphs of parabola $x = y^2$

Step 2: Find the distance between points R and A. Your answer should be the same as: $A(-p, y)$ and $R(x, y)$ So, |RA| $= \sqrt{\left(x - (-p)\right)^2 + (y - y)^2}$ $= \sqrt{(x + p)^2}$

Step 3: Find the distance between points R and F Is your result similar to the one below? If no, recalculate. $F(p, 0)$ and $R(x, y)$ So, |RF| $= \sqrt{(x - p)^2 + (y - 0)^2}$ $= \sqrt{(x - p)^2 + y^2}$

Step 4: Equate |RF| to |RA| and simplify the resulting equation. Is your result similar to the one below? If no, recalculate. $\sqrt{(x - p)^2 + y^2} = \sqrt{(x + p)^2}$ Take squares of both sides $(x - p)^2 + y^2 = (x + p)^2$ Let $x^2$ stand alone. $y^2 = (x + p)^2 - (x - p)^2$ Expand RHS using difference of two squares $y^2 = (x + p - x + p)(x + p + x - p)$ Simplify RHS $y^2 = (2p)(2x)$ $y^2 = 4px$ or $4px = y^2$

This gives the basic equation of the parabola where the directrix is parallel to the y-axis.

Example 3.10

Describe the features of the parabola $x = y^2$

Solution $x = y^2$ $4p = 1$ $p = \frac{1}{4} = 0.25$ Comparing we have $1 = \frac{1}{4p}$ …………….. equation 1 (1) to equation 2, [Figure] $4px = y^2$ making $x$ the subject, we have $x = \frac{1}{4p}y^2$ ………. (2)

This means that the focal length of this parabola is 0.25, The directrix is $x = -0.25$ Figure 3.32: Features of parabola x $= y^2$ The focus is (0.25, 0)

Example 3.11

Describe the features of the parabola $y = -x^2$ Solution have $4p = -1$ $p = -\frac{1}{4} = -0.25$ This $x = -y^2$ means $1 = -\frac{1}{4p}$ that …………….. the focal length (1) of this [Figure] $4px = y^2$ making x the subject, we have $x = -\frac{1}{4p}y^2$ ……………. (2) Comparing equation 1 to equation 2, we

parabola is 0.25, Figure 3.33: Features of parabola $x = -y^2$ The directrix is $x = 0.25$ The focus is $(-0.25, 0)$ Generally, for a parabolic equation of the form $y = a(x + h)^2 + k$ : The vertex is located at $V(-h, k)$ The Focus is located at $F\left(-h, (k + p)\right)$ The focal length is $p = \frac{1}{4a}$ .

The equation of the axis of symmetry is $x = -h$ The equation of the directrix is $y = k - p$ as illustrated in Figure 3.34. [Figure] Figure 3.34: Features of a parabola with equation $y = a(x + h)^2 + k$ Generally, for a parabolic equation of the form $x = a(y + h)^2 + k$ : The vertex is located at $V(k, -h)$ The Focus is located at $F(k + p, -h)$ The focal length is $p = \frac{1}{4a}$ . The equation of the axis of symmetry is $y = -h$ The equation of the directrix is $x = k - p$ as illustrated in Figure 3.35 [Figure] Figure 3.35: Features of a parabola with equation $x = a(y + h)^2 + k$

Example 3.12

Sketch the curve $x = 2y^2$ .

Solution

Comparing $x = 2y^2$ and $x = a(y + k)^2 + h$ , $a = 2, h = 0, k = 0$ and $p = \frac{1}{4a}$ $p = \frac{1}{4(2)}$ $p = 0.125$ The vertex is located at $(0, 0)$ The focus is located at $F(0.125, 0)$ The directrix has equation: $x = -0.125$ The axis of symmetry has equation $y = 0$ The curve opens towards the right i.e., positive $x -$ axis The curve is also obtained from a shrink of the curve of $x = y^2$ [Figure] Figure 3.36: Graph of $x = 2y^2$

Example 3.13

Sketch the curve $x = -3y^2$ .

Solution

Comparing $x = -3y^2$ and $x = A(y + B)^2 + C$ , $A = -3, B = 0, C = 0$ and $p = -\frac{1}{12}$ $p \approx -0.08$ The vertex is located at $V(0, 0)$

The focus is located at $F(-0.08, 0)$ The directrix has equation: $x = 0.08$ The axis of symmetry has equation $y = 0$ The curve opens towards the left i.e., negative $x -$ axis The curve is also obtained from a shrink of the curve of $x = y^2$ [Figure] Figure 3.37: Graph of $x = -3y^2$

Example 3.14

Sketch the curve $y = 2x^2 - 12x + 22$ .

Solution

$y = 2x^2 - 12x + 22$ $y = 2(x - 3)^2 + 4$ Comparing $y = 2(x - 3)^2 + 4$ and $y = a(x + h)^2 + k$ , $a = 2, h = -3, k = 4$ and $p \approx 0.13$ The vertex is located at $V = (-h, k) = (3, 4)$ The focus is located at $F = (-h, k + p) = (3, 4.13)$ The directrix has equation: $y = k - p = 3.87$ The axis of symmetry has equation $x = 3$ The curve opens upward The curve is also obtained from a shrink of the curve of $y = x^2$

[Figure] Figure 3.38: Graph of $y = 2x^2 - 12x + 22$

7Standard Equation of a Parabolap. 93

Earlier we derived the standard equation of a parabola with its vertex at the origin. In this lesson, we will build on that knowledge to derive the standard equation of a parabola whose vertex is located at a point other than the origin.

Deriving the standard equation of a parabola with

a vertex not at the origin

If the vertex of the parabola is not at the origin, then it is translated.
If a parabola is translated $h$ units horizontally and $k$ units vertically, the vertex will be $(h, k$ ).
This translation will cause $x$ to be replaced with $(x - h) \text{ and } y \text{ replaced with } (y - k)$ in the standard form of the equation. Recall that: A parabola is the set of all points (x, y) in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. So, for any chosen point (x, y) on the parabola, the distance between our chosen point on the parabola and the focus is equal to the perpendicular distance between our chosen point and the directrix. Let us use the activity below to derive the equation. We will first consider the situation where the directrix is parallel to the $x -$ axis

Activity 3.8 Deriving the standard equation of a parabola with directrix

parallel to the X - axis

Step 1: With the aid of a graph sheet, make a copy of the diagram below [Figure] Figure 3.39: Features of a parabola with equation $y = a(x + h)^2 + k$ In Figure 3.39, R $(x, y)$ is any point that satisfies the parabola or any point on the parabola, $(h, k)$ is the vertex or turning point, and P is the distance between the Focus and the vertex. By definition, this means that the distance between the directrix and the vertex is also P, and the equation of the directrix is $y = k - p$ . Also, a perpendicular from point R $(x, y)$ will intersect the directrix at point A $(x, k - p)$ and the coordinates of the focus will be F $(h, k + p)$ Step 2: Calculate $|RF|$ Is your result similar to the one below? If no, recalculate. $|RF| = \sqrt{(x - h)^2 + [y - (k + p)]^2}$

Step 3: Calculate $|RA|$ Is your result similar to the one below? If no, recalculate. $|RA| = \sqrt{(x - x)^2 + [y - (k - p)]^2}$

Step 4: Equate |RF| to |RA| and simplify the resulting equation.

$\Rightarrow |RF| = |RA|$ $\Rightarrow \sqrt{(x - h)^2 + [y - (k + p)]^2} = \sqrt{(x - x)^2 + [y - (k - p)]^2}$ Remove square roots by squaring both sides of the equation. $\Rightarrow (x - h)^2 + [y - k - p]^2 = (x - x)^2 + [y - k + p]^2$ Make $(x - h)^2$ stand alone $(x - h)^2 = (0)^2 + [y - k + p]^2 - [y - k - p]^2$ Use the difference of two squares strategy to evaluate the RHS of the equation $(x - h)^2 = [(y - k + p) + (y - k - p)][(y - k + p) - (y - k - p)]$ $(x - h)^2 = [2y - 2k][2p]$ $(x - h)^2 = 2(y - k) \times 2p$ $(x - h)^2 = 4p(y - k)$

Therefore, the general equation of the parabola with vertex (h, k) and focal length $\boldsymbol{p}$ is given by; $(x - h)^2 = 4p(y - k)$ When the vertex is at the origin (0, 0), the equation becomes $(x - 0)^2 = 4p(y - 0)$ which simplifies to $x^2 = 4py$ Next, let us consider the situation where the directrix is parallel to the y-axis

Activity 3.9 Deriving the standard equation of a parabola with directrix

parallel to the y - axis

Step 1: With the aid of a graph sheet, make a copy of the diagram below [Figure] Figure 3.40: Features of a parabola with equation $x = a(y + h)^2 + k$

In Figure 3.40, R $(x, y)$ is any point that satisfies the parabola or any point on the parabola? $(h, k)$ is the vertex or turning point and P is the distance between the Focus and the vertex. By definition, this means that the distance between the directrix and the vertex is also P, and the equation of the directrix is $x = h - p$ . Also, a perpendicular from point $R(x, y)$ will intersect the directrix at point A $(h - p, y)$ and the coordinates of the focus will be F $(h + p, k)$ Step 2: Calculate |RF| Is your result similar to the one below? If no, recalculate. $|RF| = \sqrt{(y - h)^2 + [x - (k + p)]^2}$

Step 3: Calculate |RA| Is your result similar to the one below? If no, recalculate. $|RA| = \sqrt{(y - y)^2 + [x - (k - p)]^2}$

Step 4: Equate |RF| to |RA| and simplify the resulting equation. $\Rightarrow |RF| = |RA|$ $\Rightarrow \sqrt{(y - k)^2 + [x - (h + p)]^2} = \sqrt{(y - y)^2 + [x - (h - p)]^2}$ Remove square roots by squaring both sides of the equation. $\Rightarrow (y - k)^2 + [x - h - p]^2 = (y - y)^2 + [x - h + p]^2$ Make $(y - k)^2$ stand alone $(y - k)^2 = (0)^2 + [x - h + p]^2 - [x - h - p]^2$ Use the difference of two squares strategy to evaluate the RHS of the equation $(y - k)^2 = [(x - h + p) + (x - h - p)][(x - h + p) - (x - h - p)]$ $(y - k)^2 = [2x - 2h][2p]$ $(y - k)^2 = 2(x - h) \times 2p$ $(y - k)^2 = 4p(x - h)$

Therefore, the general equation of the parabola with vertex (h, k) and focal length p is given by: $(y - k)^2 = 4p(x - h)$

Example 3.15

Find the equation of a parabola with a focus at $(1, 2)$ and vertex at $(-2, 2).$

Solution

Vertex $= (-2, 2)$ Focus $=$ $(1, 2)$ The y-coordinates of the vertex and the focus are the same. This shows that the line of symmetry is $y = 2$ Also, the equation of the parabola is given by $(y - k)^2 = 4p(x - h)$ Substituting the vertex, we have $(y - 2)^2 = 4p(x - (-2))$ $\Longrightarrow (y - 2)^2 = 4p(x + 2)$ The focal length, $\boldsymbol{p}$ , is the distance between the focus and the vertex. $p = \sqrt{(2 - 2)^2 + (-2 - 1)^2}$ $= \sqrt{3^2}$ $= 3$ The required equation is $(y - 2)^2 = 4 \times 3(x + 2)$ $(y - 2)^2 = 12(x + 2)$

Example 3.16

Find the equation of a parabola with a focus at $(4, -3)$ and vertex at $(4, 2).$

Solution

Vertex $= (4, 2)$ Focus $=$ $(4, -3)$ The x-coordinates of the vertex and the focus are the same. This shows that the line of symmetry is $x = 4$ Also, the equation of the parabola is given by $(x - h)^2 = 4p(y - k)$ substituting the vertex, we have $(x - 4)^2 = 4p(y - 2)$ $\Longrightarrow (x - 4)^2 = 4p(y - 2)$ The focal length, $\boldsymbol{p}$ , is the distance between the focus and the vertex. $p = \sqrt{(-3 - 2)^2 + (4 - 4)^2}$ $= \sqrt{(-5)^2}$ $= \sqrt{-5 \times -5} = \sqrt{25} = 5$ The required equation is $(x - 4)^2 = 4 \times 5(y - 2)$ $(x - 4)^2 = 20(y - 2)$

8Deducing the Directrix and Focus for a Parabolic Equation and Vice Versap. 98

A parabola can be described algebraically or geometrically. Algebraically, it is written either as: $y = a(x - h)^2 + k$ (opens up or down) OR $x = a(y - k)^2 + h$ (opens right or left) Geometrically, it is written as: $(x - h)^2 = 4p(y - k)$ (opens up or down) OR $(y - h)^2 = 4p(x - k)$ (opens left or right) Understanding the relationship between its equation, focus, and directrix is essential to analysing and graphing parabolas.

How to Deduce the Focus and Directrix from the

Equation

Step 1: Compare it to the standard equation Step 2: Deduce the focus and the directrix from the standard equation. • If it is of the form $(x - h)^2 = 4p(y - k)$ , then the Focus: $(h, k + p)$ , Directrix: $y = k - p$ , and Focal length: ∣ p ∣ • If it is of the form $(y - k)^2 = 4p(x - h)$ , then the Focus: $(h + p, k)$ , Directrix: $x = h - p$ and Focal length: ∣ p ∣

Example 3.17

Find the focus and directrix of $y^2 = 12x$ . Solution Compare the equation given to the standard equation of a parabola $(y - k)^2 = 4p(x - h)$ , $y^2 = 12x$ (Write in standard form) $(y - 0)^2 = 12(x - 0)$ , $\therefore 4p = 12$ $p = 3$ So, we have the vertex $(h, k) = (0, 0)$ Therefore, the focus of the parabola is $(0 + 3, 0) = (3, 0),$ and the directrix is $x = 0 - 3 = -3$ .

Example 3.18

Find the directrix and focus of $(x + 1)^2 = -8(y - 4)$ .

Solution

$(x + 1)^2 = -8(y - 4)$ is of the form $(x - h)^2 = 4p(y - k)$ , Comparing $(x + 1)^2 = -8(y - 4)$ to $(x - h)^2 = 4p(y - k)$ , we have the vertex of the parabola (h, k) $= (-1, 4)$ Also $4p = -8$ , and $p = -2$ Since $p < 0$ , the parabola opens down, $\therefore \text{the cordinates of the focus is } (\mathrm{h}, \mathrm{k} + \mathrm{p})$ $\therefore \text{focus} = (-1, 4 + (-2))$ $= (-1, 2)$ Directrix is $y = 4 - (-2)$ , so in this case it is $y = 6$

9Describing the Orientation of a Parabola with the Given Equationp. 99

At the beginning of this section, we explored the four orientations of a parabola. We learned how to find the equation of a parabola given its focus and directrix. Now we will apply everything we've learned to sketch parabolas. The orientation of a parabola is the direction in which the parabola opens on the coordinate plane. Knowing the orientation will help sketch the parabola. $(x - h)^2 = 4p(y - k)$ Opens 1. Vertical up if Parabola $a > 0$ or (opens $p > 0$ up [Figure] or down) $y = a(x - h)^2 + k$ or

Opens down if $a < 0$ or $p < 0$ Figure 3.41: Graph showing orientation of parabola $y = a(x - h)^2 + k$

Opens Opens Horizontal right left if if $a < 0$ Parabola $a > 0$ or or $p < 0$ $p > 0$ (opens [Figure] left or right): $x = a(y - k)^2 + h$ or $(y - k)^2 = 4p(x - h)$

Figure 3.42: Graph showing orientation of parabola $x = a(y - h)^2 + k$

How to Describe the Orientation:

Step 1: Identify which variable is squared. If $x$ is squared, it implies the parabola opens up or down If $y$ is squared, it implies the parabola opens left or right Step 2: Check the sign of the coefficient of the squared term ( $a$ or $p$ ) If a or p is positive, the parabola opens up (if vertical) or right (if horizontal) If a or p is negative, the parabola opens down (if vertical) or left (if horizontal)

Example 3.19

Describe the orientation of the parabola with the equation $(y + 1)^2 = -5(x - 3)$ . Give the parabola’s vertex and focus.

Solution

since y is squared, the parabola either opens to the right or the left. It has a horizontal line of symmetry. Comparing $(y + 1)^2 = -5(x - 3)$ with $(y - k)^2 = 4p(x - h)$ , we have $4p = -5$ $p = -\frac{5}{4} = -1.25$ Since P is negative, it means the parabola opens to the left. The vertex is $(3, -1)$ and the focus is $(1.75, -1)$

Example 3.20

Describe the orientation of the parabola with the equation $(x - 5)^2 = 4(y - 2)$ . Give the parabola’s vertex and focus.

Solution

Since $x$ is squared, the parabola opens up or down. It has a vertical line of symmetry. $(x - 5)^2 = 4(y - 2)$ The vertex is $(5, 2).$ $4p = 4$ $p = 1$ Since P is positive, the parabola opens up. The focus of the parabola is $(5, 3)$ .

10Sketching Parabolas Given the Directrix and Focusp. 101

Step 1: Plot the focus and draw the directrix in the x-y plane or a coordinate grid. Step 2: Find the vertex, which is halfway between the focus and the directrix. Step 3: Determine the orientation. If the directrix is horizontal, the focus will open to the left or right. If the directrix is vertical, the parabola will open up or down. Step 4: Sketch the parabola. The turning point is the vertex. Draw a smooth curve that passes through the vertex, curves around the focus, and moves away from the directrix.

Example 3.21

A parabola has a focus at $(-1, 2)$ and directrix $x = 1$ . a. Sketch the parabola b. Write the equation of the parabola

Solution

a. Step 1: Plot the focus $(-1, 2)$ and draw the directrix $x = 1$

[Figure] Figure 3.43: Graph $x = 1$ and point $F(-1, 2)$

Step 2: From Figure 3.43, the vertex will be located at (0, 2) as it is equidistant from the focus and the directrix. Step 3: Since the parabola has to pass around the focus, it will open to the left. Step 4: Draw a curve that opens to the left and passes through the vertex, (0, 2). Note that figure is drawn to scale. Your sketch should not necessarily be drawn to scale. [Figure] Figure 3.44: Parabola with focus at $(3, -3)$ and directrix $x = 4$

b. From Figure 3.44, the vertex is (0,2) and the focus is (-1, 2). The focal length, p, is the distance between the vertex and the focus. From the graph, p=1. You can also calculate it using the distance formula $|p| = \sqrt{(2 - 2)^2 + (-1 - 0)^2} = \sqrt{0 + 1} = \sqrt{1} = 1$

Since the parabola opens to the left, $p = -1$ Using $(y - k)^2 = 4p(x - h)$ by substituting (h, k)=(0, 2) and p=-1, we have: $(y - 2)^2 = 4(-1)(x - 0)$ $(y - 2)^2 = -4x$

Example 3.22

A parabola has a focus at $(3, 1)$ and directrix $y + 3 = 0$ . a. Make a sketch of the parabola b. Write the equation of the parabola

Solution

a. Step 1: plot the focus (3, 1) and draw the directrix $y = -3$ [Figure] Figure 3.45: Graph $y = -3$ and point F $(3, 1)$ Step 2: From Figure 3.45, the vertex will be located at $(3, -1)$ as it is equidistant from the focus and the directrix. Step 3: Since the parabola has to pass around the focus, it will open upwards. Step 4: Draw a curve that opens upwards and passes through the vertex, $(3, -1).$ [Figure] Figure 3.46: Parabola with focus at $(3, 4)$ and directrix $y = -3$

b. From the graph, the vertex is $(3, -1$ ) and the focus is $(3, 1)$ . The focal length, $p$ , is the distance between the vertex and the focus. From the graph, |p|=2. You can also calculate it using the distance formula $|p| = \sqrt{\left(1 - (-1)\right)^2 + (3 - 3)^2} = \sqrt{2^2} = 2$ Since the parabola opens upwards, $p = 2$ Using $(x - h)^2 = 4p(y - k)$ by substituting $(h, k) = (3, -1)$ and $p = 2$ , we have: $(x - 3)^2 = 4(2)\left(y - (-1)\right)$ $(x - 3)^2 = 8(y + 1)$

11Describing the Equations of Lines and Derivatives of Functionsp. 104

An equation of a straight line shows the relationship between $x$ and $y$ coordinates on a graph. In slope-intercept form, it is given by: $y = mx + c$ , where $m$ is the gradient and $c$ is the y-intercept. The gradient is calculated using the formula, $m = \frac{\Delta y}{\Delta x}$ or $m = \frac{y_2 - y_1}{}$ or $m = \frac{y_1 - y_2}{x_1 - x_2}$ Description of a line equation

If $m > 0$ , the graph strictly increases. The line slopes upwards from left to right.
when $m < 0$ , it strictly decreases. The line slopes downwards from left to right
When $m = 0$ , the line is horizontal or parallel to the x-axis. The equation of the line takes the form $y = k$ , where $k$ is a constant.
When $m = \infty$ (undefined), the line equation is vertical or parallel to the y-axis. The equation of the line takes the form $x = k$ , where $k$ is a constant [Figure] Figure 3.47: Graphs showing description of line equation. The gradient of a function can also be obtained using the gradient function. The first derivative of a function gives the gradient function. For parabolas (quadratic functions), the derivative will be a linear equation. The gradient of the parabola changes depending on the point of contact of the parabola and the line.

12Determining and Describing the Equations of Tangents and Normal to a Parabolap. 105

Activity 3.10 Finding the equation of a tangent or normal to a parabola

Step 1: Locate the point at which the line intersects the parabola, $(x_1, y_1)$ . If one of the coordinates of the points is provided, substitute it into the equation of the parabola and find the other coordinates. It is possible to have multiple points. Step 2: Find the first derivative, $\left(\frac{dy}{dx}\right)$ , of the parabola. Step 3 : Substitute $(x_1, y_1)$ into the derivative function. This gives the gradient of the tangent. gradient $(m)$ $= \left.\frac{dy}{dx}\right|_{(x_1, y_1)}$ The gradient of the normal $= -\frac{1}{m}$ Step 4 : Use $(x_1, y_1)$ and the gradient to find the tangent and/or normal equation. The slope of a tangent is given by $\frac{dy}{dx}$ . The slope of the normal is given by $-\frac{dx}{dy} = -\frac{1}{\frac{dy}{dx}}$ Let us now generate an equation of a tangent to a parabola of the form $y^2 = 4px$ Differentiating $y^2 = 4px$ on both sides with respect to $x$ will be $2y\frac{dy}{dx} = 4p$ $\frac{dy}{dx} = \frac{2p}{y}$ Therefore, the slope $(m)$ of the tangent to the parabola at the point $(x_1, y_1)$ on it is $\frac{2p}{y_1}$ . Now, the equation of the tangent to the parabola at the point $(x_1, y_1)$ on it is $y - y_1 = m(x - x_1)$ $\Rightarrow y - y_1 = \frac{2p}{y_1}(x - x_1)$ $\Longrightarrow yy_1 - y_1^2 = 2px - 2px_1$ But $y_1^2 = 4px$ $yy_1 - 4px_1 = 2px - 2px_1$

$\therefore yy_1 = 2p(x + x_1)$ is the equation of the tangent at a point $(x_1, y_1)$ on the parabola. For a parabola with equation of the form $x^2 = 4py$ , Differentiating $x^2 = 4py$ on both sides with respect to $x$ will be $2x = 4p\frac{dy}{dx}$ $\frac{dy}{dx} = \frac{2x}{4p} = \frac{x}{2p}$ Therefore, the slope $(m)$ of the tangent to the parabola at the point $(x_1, y_1)$ on it is $\frac{x_1}{2p}$ . Now, the equation of the tangent to the parabola at the point $(x_1, y_1)$ on it is $y - y_1 = m(x - x_1)$ $\Rightarrow y - y_1 = \frac{x_1}{2p}(x - x_1)$ $\Longrightarrow xx_1 - x_1^2 = 2py - 2py_1$ But $x_1^2 = 4py_1$ $\Rightarrow xx_1 - 4py_1 = 2py - 2py_1$ $\therefore 2py = xx_1 - 2p(2y_1 - y_1)$ is the equation of the tangent at a point $(x_1, y_1)$ on the parabola. Equation of a normal to a parabola of the form $y^2 = 4px$ Recall that the relationship between the gradients of two lines which a perpendicular, in this case, a tangent and a normal, given that their gradients are $m_1$ and $m_2$ is $m_1 \times m_2 = -1$ or $m_1 = -\frac{1}{m_2}$ Since the gradient of the tangent to the parabola with equation $y^2 = 4px$ at $(x_1, y_1)$ is $\frac{2p}{y_1}$ , the slope $(m)$ of the normal is $-\frac{y_1}{2a}$ since normal and tangent are perpendicular to each other. $\therefore$ The equation of the normal at a point $(x_1, y_1)$ on the parabola becomes $y - y_1 = -\frac{y_1}{2p}(x - x_1)$ $\Longrightarrow y_1x - 2py_1 - x_1y_1 + 2py = 0$ Similarly, the slope $(m)$ of the normal to the parabola with the equation $x^2 = 4py$ at $(x_1, y_1)$ is $-\frac{2p}{x_1}$ Since the gradient of the tangent is $\frac{x_1}{2p}$ . $\therefore$ The equation of the normal at a point $(x_1, y_1)$ on the parabola becomes

$y - y_1 = -\frac{2p}{x_1}(x - x_1)$ $\Longrightarrow x_1y + 2px - x_1y_1 - 2px_1 = 0$

Example 3.23

A parabola $y^2 = 8x$ passes through the point $P(2, 4)$ . Find: a. Equation of the tangent to the parabola at the point $P$ b. Equation of the normal to the parabola at the point $P$

Solution

If $y^2 = 8x$ , then comparing it to $y^2 = 4px, \ p = 2$ a. The equation of the tangent to the parabola is given as $yy_1 = 2p(x + x_1)$ $\Rightarrow y(4) = 2 \times 2(x + 2)$ $y = x + 2$ Alternatively, $y^2 = 8x$ $\Rightarrow 2y\frac{dy}{dx} = 8$ $\frac{dy}{dx} = \frac{4}{y}$ The gradient of the tangent line at $P$ , $\left.\frac{dy}{dx}\right|_{y=4} = \frac{4}{4} = 1$ The equation of the tangent line that passes through $P$ $y - 4 = 1(x - 2)$ $y = x + 2$ b. The equation of the normal is given by $y_1x - 2py_1 - x_1y_1 + 2py = 0$ $\Longrightarrow 4x - 2(2)4 - (2)(4) + 2(2)y = 0$ $y = -x + 6$ Alternatively, The gradient of the normal line at $P = -\frac{1}{1} = -1$ The equation of the tangent line that passes through $P$ $y - 4 = -1(x - 2)$ $y = -x + 6$

Example 3.24

The curve is defined by the equation $(y - 3)^2 = -4x$ (a) Find the equation of the tangent to the curve at point ( $-4, -1$ ). (b) Find the equation of the normal to the curve at point ( $-4, 7$ ).

Solution

(a)The gradient of the tangent is $\frac{dy}{dx}$ $\frac{d}{dx}\left((y - 3)^2 = -4x\right)$ $2(y - 3)\frac{dy}{dx} = -4$ $\frac{dy}{dx} = -\frac{4}{2(y - 3)} = -\frac{2}{y - 3}$ $= \left.\frac{dy}{dx}\right|_{(-4,-1)} = -\frac{2}{-1 - 3} = \frac{1}{2}$ Equation of the tangent with gradient $\frac{1}{2}$ passing through $(-4, -1)$ is: $y - (-1) = \frac{1}{2}\left(x - (-4)\right)$ $y + 1 = \frac{1}{2}(x + 4)$ $y = \frac{1}{2}x + 1$ or $2y - x - 2 = 0$ (b) At point $(-4, 7)$ , the gradient of the tangent is $= -\frac{2}{7 - 3} = -\frac{2}{4} = -\frac{1}{2}$ The gradient of the normal is $-1 \div -\frac{1}{2} = -1 \times -\frac{2}{1} = 2$ So, the equation of the normal at $(-4, 7)$ is: $y - 7 = 2\left(x - (-4)\right)$ $y - 7 = 2(x + 4)$ $y = 2x + 8 + 7$ $y = 2x + 15$

Example 3.25

A water jet follows the parabolic path $y = 14x - 2x^2$ direction Determine the A sensor sensor's of is the placed the line. point water along where at $x = 2$ the the path, . jet Find reaches perpendicular the equation its highest to of the [Figure] point. Find the gradient of the tangent at $x = 2$ .

Figure 3.48: Water jet

Solution

a. For maximum height, put $\frac{dy}{dx} = 0$ $\frac{dy}{dx} = 14 - 4x$ $\Longrightarrow 14 - 4x = 0$ $x = -\frac{14}{-4}$ $= 3.5$ When $x = 3.5$ , $y = 14 \times 3.5 - 2(3.5)^2$ $y = 24.5$ units. The maximum point reached by the water jet is ( $3.5, 24.5$ ) b. The gradient of the tangent at $x = 2$ is $14 - 4(2) = 14 - 8 = 6$ c. When $x = 2, y = 14(2) - 2(2)^2 = 28 - 8 = 20$ The gradient of the path of the sensor is the same as the gradient of the normal to the curve $x = 2$ Therefore, the gradient of the path of the sensor $= -\frac{1}{6}$ The equation of the path of the sensor is; $y - 20 = -\frac{1}{6}(x - 2)$ $y = -\frac{1}{6}x + \frac{1}{3} + 20$ $y = -\frac{1}{6}x + 20\frac{1}{3}$ Hence the equation of the sensor is $y = -\frac{1}{6}x + \frac{61}{3}$ or $6y + x - 122 = 0$

13Finding the Point of Intersection Between a Line and a Parabolap. 109

A straight line will intersect a parabola at a maximum of two points. Note that a line that intersects a curve at two or more points is called a secant. If the line intersects the parabola at one point, the line might be tangent to the parabola. [Figure] [Figure] [Figure] Figure 3.49: Line intersects Figure 3.50: Line intersects Figure 3.51: Line does not parabola at two points, A and B parabola at one point, A. intersect parabola

How to find the point of intersection of a line and a

parabola

Step 1: Set the equations equal. You might also make a variable in the linear equation the subject and substitute the result into the parabolic equation. Step 2: Rearrange your result from step 1 to form a quadratic equation Step 3: Solve the quadratic equation to find one of the variables. [If the discriminant of the resulting quadratic is 0, you can be sure the line is tangent to the parabola. If the discriminant is less than 0, the line and the parabola do not intersect. In such cases, step 4 will not be necessary.] Step 4: Using your result from step 3, find the corresponding x or y-values.

Example 3.26

The line $2y + 3x - 6 = 0$ intersects the parabola $y^2 = 6x$ at $K$ and $M$ respectively. a. Find the coordinates of $K$ and $M$ . b. Find the equation of the normal to the parabola at the point $\left(\frac{2}{3}, 2\right)$ .

Solution

a. Step 1: Set the two equations equal b. Let $2y + 3x - 6 = 0 \ldots\ldots (1)$ and $y^2 = 6x \ldots\ldots\ldots\ldots\ldots (2)$ From (1), we have the line equation: $x = \frac{6 - 2y}{3}$ From (2), we have the parabolic equation: $x = \frac{1}{6}y^2$ Since the line and parabola intersect, $\frac{6 - 2y}{3} = \frac{y^2}{6}$ Step 2: Rearrange $\frac{6 - 2y}{3} = \frac{y^2}{6}$ to form a quadratic equation. multiply through by 6 $6 \times \left(\frac{6 - 2y}{3}\right) = 6 \times \left(\frac{y^2}{6}\right)$ $2(6 - 2y) = y^2$ $12 - 4y - y^2 = 0$ multiply through by $-1$

$y^2 + 4y - 12 = 0$ Step 3: Solve the resulting quadratic equation $y^2 + 6y - 2y - 12 = 0$ $y(y + 6) - 2(y + 6) = 0$ $(y - 2)(y + 6) = 0$ This means that either $(y - 2) = 0 \Longrightarrow y = 2$ or $y + 6 = 0 \Longrightarrow y = -6$ $y = 2$ or $y = -6$ Step 4: For each y-value, find its corresponding x-value When $y = 2, \ x = \frac{1}{6}(2)^2 = \frac{2}{3}$ and when $y = -6, x = \frac{1}{6}(-6)^2 = \frac{1}{6} \times 36 = 6$ $\therefore K = \left(\frac{2}{3}, 2\right)$ and $M = (6, -6)$ c. The gradient function of the parabola is $\frac{d}{dx}(y^2 = 6x)$ $2y\frac{dy}{dx} = 6$ $m = \frac{dy}{dx} = \frac{6}{2y} = \frac{3}{y} = \frac{3}{2}$ The gradient of the normal is $= -1 \div \frac{3}{2} = -\frac{2}{3}$ The equation of the normal to the parabola at $\left(\frac{2}{3}, 2\right)$ is: $y - 2 = -\frac{2}{3}\left(x - \frac{2}{3}\right)$ $y - 2 = -\frac{2}{3}x + \frac{4}{9}$ $9y - 18 = -6x + 4$ $9y + 6x - 22 = 0$ Alternatively, Using the equation $2py - 2py_1 + xy_1 - x_1y_1 = 0$ We have $y^2 = 6x$ and $p = \frac{3}{2}$ The point is $\left(\frac{2}{3}, 2\right)$ therefore, the equation of the normal is $2\left(\frac{3}{2}\right)y - 2\left(\frac{3}{2}\right) \cdot 2 + (2)x - 2 \cdot \left(\frac{2}{3}\right) = 0$ $9y + 6x - 22 = 0$

Example 3.27

cable The $y - 3x - 4 = 0$ arch is suspended of a bridge, straight built in along the shape the line of [Figure] a parabola, is given by $y = 8x - x^2$ . A

Find the point(s) where the cable touches Figure 3.52: A bridge the bridge. State whether the cable is a tangent or a secant to the bridge.

Solution

a. We will follow the steps to answer the questions Step 1: Set the two equations equal Parabolic equation: $y = 8x - x^2$ . Line equation: $y = 3x + 4$ Setting the equations equal, we have: $8x - x^2 = 3x + 4$ Step 2: Rearrange $8x - x^2 = 3x + 4$ to form a quadratic equation. $8x - x^2 - 3x - 4 = 0$ $-x^2 + 5x - 4 = 0$ (multiply through by -1) $x^2 - 5x + 4 = 0$ Step 3: Solve $x^2 - 5x + 4 = 0$ for the values of $x$ $x^2 - 5x + 4 = 0$ $x^2 - x - 4x + 4 = 0$ $x(x - 1) - 4(x - 1) = 0$ $(x - 1)(x - 4) = 0$ So, $x - 1 = 0$ and/or $x - 4 = 0$ This implies that $x = 1$ and/or $x = 4$ Step 4: For each x-value, find its corresponding y-value. If $x = 1$ , $y = 3(1) + 4 = 7$ If $x = 4$ , $y = 3(4) + 4 = 16$ So, the cable touches the bridge at (1, 7) and (4, 16). b. The cable is secant to the arch of the bridge since it intersects the bridge at two points.

Example 3.28

A parabola has a focus at $(0, -3)$ and directrix $x - 2 = 0$ . a. Write the equation of the parabola b. Find the point at which the parabola intersects the line $2y + x + 1 = 0$ and state the relationship between the parabola and the line. c. Find the equation of the tangent to the parabola at $y = -5$

Solution

a. Therefore, $(y + 3)^2 = 4(-1)(x - 1)$ A sketch $(y + 3)^2 = -4(x - 1)$ of the the equation parabola of shows the parabola that the is vertex given is by: [Figure] $(1, -3)$ and the focal length is $|p| = 1$ Since the parabola opens to the left, it means $p = -1.$

Figure 3.53: Parabola with focus $(0, -3)$ and directrix $x = 2$ b. The equation of the parabola is: $(y + 3)^2 = -4(x - 1)$ The line equation is: $2y + x + 1 = 0$ From the line equation, we have, $x = -2y - 1$ Substituting $x = 2y - 1$ into the parabolic equation, we have: $(y + 3)^2 = -4((-2y - 1) - 1)$ Simplify and solve the resulting quadratic equation. $(y + 3)^2 = -4(-2y - 2)$ $y^2 + 6y + 9 = 8y + 8$ $y^2 - 2y + 1 = 0$ $y^2 - y - y + 1 = 0$ $y(y - 1) - 1(y - 1) = 0$ $(y - 1)(y - 1) = 0$ $y - 1 = 0$ $\Longrightarrow y = 1$ If $y = 1, x = -2(1) - 1 = -3$ So, the point of intersection of the parabola and the line is ( $-3, 1)$ . Since the line intersects the parabola at one point, the line is tangent to the parabola at $(-3, 1)$

c. We will first find the value of x for which $y = -5$ . To do this, we will substitute $y = -5$ into the parabola and solve for x $(-5 + 3)^2 = -4(x - 1)$ $4 = -4(x - 1)$ divide each side by $-4$ $\frac{4}{-4} = -\frac{4}{-4}(x - 1)$ $-1 = x - 1$ $-1 + 1 = x$ $0 = x$ This means that the tangent intersects the parabola at $(0, -5)$ Next, we find the gradient: $\frac{d}{dx}[(y + 3)^2 = -4(x - 1)]$ $2(y + 3)\frac{dy}{dx} = -4$ $\frac{dy}{dx} = -\frac{4}{2(y + 3)}$ At $(0, -5),$ the gradient of the tangent is $= \frac{dy}{dx} = -\frac{4}{2(-5 + 3)} = -\frac{4}{-4} = 1$ Therefore, the equation of the tangent is: $y - (-5) = 1(x - 0)$ $y + 5 = x$ $y - x + 5 = 0$

14Review questionsp. 115

Outline 5 real-life examples of parabolas.
In each of the following graphs, a. Describe the shape. Your description should include a statement about the domain, range, line of symmetry, orientation, and maximum or minimum value b. Write the quadratic function in the form $y = a(x + h)^2 + k$ where a, h, and k are parameters. c. Describe the shape of the graph as compared to $y = x^2$ (i) [Figure] (ii) [Figure] Figure 3.54: Parabola 1 Figure 3.55: Parabola 2 (iii) [Figure] (iv) [Figure] Figure 3.56: Parabola 3 Figure 3.57: Parabola 4 (v) [Figure] (vi) [Figure] Figure 3.58: Parabola 5 Figure 3.59: Parabola 6
The parabola is defined by: $y(y - 2) = 4x - 3$ .

a. Sketch the parabola. Your sketch should include the coordinates of the focus, the coordinates of the vertex, the line of symmetry, and the points at which the parabola intersects with the x-axis and the y-axis. b. Give a brief description of the orientation of the parabola. 4. The graph below shows a cross-section of a drainage in the form of a parabola. [Figure] Figure 3.60: Parabola representing cross-section of a drainage a. Write the equation of the parabola in the form $y = a(x - h)^2 + k$ where a, h, and k are parameters. b. Describe the shape of the graph as compared to $y = x^2$ c. State the focus, the equation of the directrix, and the focal length. d. Find the points of intersection of the parabola and the line 7y+4x-4=0 5. Describe and sketch the graph of $y = -x^2 + 3$ . 6. Sketch the graph $y = 3 - 5x - 2x^2$ and compare your graph to $y = x^2$ . 7. Sketch the graph $y = x^2 - 7x + 6$ and compare your graph to $y = x^2$ . 8. The diagram below shows a graph of a parabola. Point V is the vertex.

[Figure] Figure 3.61: Graph of a parabola

a. Write the equation of the parabola. b. Describe the shape of the graph as compared to $x = y^2$ 9. Figure 3.62 shows a graph of a quadratic function. [Figure] Figure 3.62: Graph of a parabola a. Write the quadratic function in the form $x = a(y - h)^2 + k$ where a, h, and k are parameters. b. Describe the shape of the graph when compared to $x = y^2$ 10. The graph below represents the path of a volleyed ball, where H is the height of the ball in meters, and t is the time in seconds after it was hit.

[Figure] Figure 3.63: Graph of a parabola representing the path of a volleyball

a. Write a quadratic equation to represent the path of the projectile. b. Describe the shape of the H in Figure 3.63 11. Sketch and describe the features of the parabola $x + 12y^2 = 0$ .

A parabola has a focus at $(-4, -2)$ and vertex at $(-2, -2).$ a. Find the equation of the parabola b. Sketch the parabola. Your sketch should include the directrix, the coordinate of the x-axis, focus and vertex.
A parabola has a focus at $(-1, 3)$ and vertex at $(2, 3).$ a. Find the equation of the parabola. b. Sketch the parabola. Your sketch should include the directrix, the coordinate of the x-axis, focus and vertex.
Find the vertex, focus and directrix of the following parabolas: a. $y^2 = -7x$ b. $(x - 2)^2 = -8(y - 5)$ c. $(y - 1)^2 = -8(x + 4)$
Describe the orientation of the parabola with the following equations equation a. $y^2 = -8x$ b. $(x - 2)^2 = -8(y - 5)$ c. $(y - 1)^2 = -8(x + 4)$ .
A parabola has a focus at $(-3, 3)$ and directrix $x = 0$ . a. Sketch the parabola. Your sketch should include the point of intersection with the $x -$ axis, the coordinates of the vertex. b. Write the equation of the parabola
A parabola has a focus at $(2, -1)$ and directrix $y + 4 = 0$ . a. Sketch the parabola. Your sketch should include the points of intersection with the coordinates, the coordinates of the vertex. b. Write the equation of the parabola
A Parabola $y^2 = 3x$ passes through the point $P(3, 3)$ . a. Find the equation of the tangent to the parabola at P b. Find the equation of the normal to the parabola at the point P. c. The parabola intersects $2y - 3x + 3 = 0$ at points P and Q. Find i. the coordinates of Q ii. PQ, leave your answer in surds iii. the equation of the tangent to the parabola at point Q.
The curve is defined by the equation: $(y - 4)^2 = -8x$ a. Find the equation of the tangent to the curve at the point where the parabola intersects the $x -$ axis.

b. Find the equation of the normal to the curve at $y = 8.$ c. Find the point at which the parabola intersects $y - 3x - 4 = 0$ 20. A parabola has a vertex at $(-1, -1)$ and directrix $2y - 1 = 0$ . a. Write the equation of the parabola b. Find the point at which the parabola intersects the line $3y + 2x + 5 = 0$ and state the relationship between the parabola and the line. c. Find the equation of the tangent to the parabola at its y-intercept. d. Find the equation of the normal to the parabola at $x = -4$

Study

KEY IDEAS

Activity 3.1 Discussing Real-life examples of Parabolas

Shape of a Quadratic Function

Maximum graph

Minimum graph

Example 3.1

Solution

Example 3.2

Solution

Activity 3.2 Graph of the quadratic equation when IaI>1, h = 0 and k = 0

Activity 3.3 The graph of the quadratic equation when -1 < a < 1, h = 0 and

k = 0

Activity 3.4 Graph of the quadratic function when h is any number other

than 0

General observation about the value of h

Activity 3.5 Shape of parabolic graphs when the value of “ k\boldsymbol{k}k ” is not 0

General observation about the value of k

Example 3.3

Solution

Example 3.4

Solution

Example 3.5

Example 3.6

Solution

Example 3.7

Solution

Definition of Parabola

The Focus

Directrix of a Parabola

Line of symmetry of a Parabola

Focal Length of a Parabola (p)

Vertex

Opening direction

Activity 3.6 Generating the parabolic equation with vertex (0, 0), focal

length p, and a directrix which is parallel to the x-axis.

Example 3.8

Example 3.9

Activity 3.7 Generating the parabolic equation with vertex (0, 0), focal

length P, and a directrix which is parallel to the y-axis.

Example 3.10

Example 3.11

Example 3.12

Solution

Example 3.13

Solution

Example 3.14

Solution

Deriving the standard equation of a parabola with

a vertex not at the origin

Activity 3.8 Deriving the standard equation of a parabola with directrix

parallel to the X - axis

Activity 3.9 Deriving the standard equation of a parabola with directrix

parallel to the y - axis

Example 3.15

Solution

Example 3.16

Solution

How to Deduce the Focus and Directrix from the

Equation

Example 3.17

Example 3.18

Solution

How to Describe the Orientation:

Example 3.19

Solution

Example 3.20

Solution

Example 3.21

Solution

Example 3.22

Solution

Activity 3.10 Finding the equation of a tangent or normal to a parabola

Example 3.23

Solution

Example 3.24

Solution

Example 3.25

Solution

How to find the point of intersection of a line and a

Activity 3.5 Shape of parabolic graphs when the value of “ $\boldsymbol{k}$ ” is not 0