HMH Algebra 2, Volume 1 1st Edition Module 4 Chapter 4 Exercise 4.2 Quadratic Equations And System Of Equations

Algebra 2 Volume 1 1st Edition Module 4 Quadratic Equations And System Of Equations

Page 175 Problem 1 Answer

A statement is given.

It is required to explain how distance formula is connected with deriving the equations for horizontal and vertical parabolas.

A parabola is a locus of a point which is equidistant from a fixed line and a fixed point.

To find the equation of a parabola, the distance between a point and the fixed point is equated to the distance between the point and the fixed line.

Therefore, the distance formula is needed in deriving the equation of a parabola.

The distance formula is connected with deriving the equations for horizontal and vertical parabolas as the distance between a point and the fixed point is equated to the distance between the point and the fixed line.

Page 175 Problem 2 Answer

An incomplete statement is given as the coordinates of the focus of the parabola are given by ____.

It is required to complete the statement.

The coordinates of the focus of the parabola are given by (p,0).

The complete statement is the coordinates of the focus of the parabola are given by (p,0).

Page 175 Problem 3 Answer

A statement is given.

It is required to write the expression for the distance from a point (x,y) on the parabola to the focus of the parabola.

The coordinates of the focus of a parabola is given as (p,0).

The distance formula between two points (x₁,y₁),(x₂,y₂) is given as d=√(x₁−x₂)²+(y₁−y₂)².

Therefore, the distance from a point (x,y) on the parabola to the focus of the parabola is d=√(x−p)²+y².

The distance from a point (x,y) on the parabola to the focus of the parabola is d=√(x−p)²+y².

Page 175 Problem 4 Answer

It is given that the equation of the directrix of a parabola is given as x=−p.

It is required to find the point of intersection of a horizontal line from a point (x,y) on the parabola and the directrix of the parabola.

The x−coordinate of every point on the directrix is −p.

Therefore, the point of intersection of a horizontal line and the directrix is (−p,y).

The point of intersection of a horizontal line from a point (x,y)

on the parabola and the directrix of the parabola is (−p,y).

Page 175 Problem 5 Answer

It is given that a horizontal line from a point (x,y) on the parabola intersects the directrix of the parabola.

It is required to write the expression for the distance between the point on the parabola and the point of intersection between the horizontal line and the directrix.

From part (c) of the exercise, the point of intersection between the horizontal line and the directrix is (−p,y).

The distance formula between two points (x₁,y₁),(x₂,y₂) is given as d=√(x¹−x²)²+(y₁−y₂)².

Therefore, the distance from a point (x,y) on the parabola to the mentioned point of intersection is d=√(x−(−p))²+(y−y)².

On simplifying, d=x+p.

The distance between the point on the parabola and the point of intersection between the horizontal line and the directrix is d=x+p.

Page 175 Problem 6 Answer

A statement is given.

It is required to equate the distance between a point on the parabola and the focus to the distance between the point and the directrix and simplify the equation.

From part (b) of the exercise, the distance from a point (x,y) on the parabola to the focus of the parabola is d=√(x−p)²+y².

From part (d) of the exercise, the distance between the point on the parabola and the point of intersection between the horizontal line and the directrix is d=x+p.

Equate both the distances and square on both sides of the equation. Expand the expression by using the formula (a+b)²=a²+2ab+b².

Equate both the distances and square on both sides of the equation. Expand the expression by using the formula (a+b)2=a²+2ab+b².

(√(x−p)²+y2)²=(x+p)²

(x−p)²+y²=x²+2px+p²

x²−2px+p²+y²=x²+2px+p²

The equation on equating the distance between a point on the parabola and the focus to the distance between the point and the directrix is x²−2px+p²+y²=x²+2px+p².

Page 175 Problem 7 Answer

A statement is given.

It is required to collect the terms in the equation obtain in part (e) of the exercise.

From part (e) of the exercise, the equation obtained is x²−2px+p²+y²=x²+2px+p².

On collecting the like terms, the equation simplifies to (x²−x²)+(−2p−2p)x+(p²−p²)+y²=0.

On simplifying further, 0.x²−4px+0.p²+y²=0.

The simplified equation of a parabola is 0.x²−4px+0.p²+y²=0.

Page 175 Problem 8 Answer

It is given that 0⋅x²−4px+0⋅p²+y²=0.

It is required to convert this in the horizontal form of parabola.

The given equation is 0⋅x²−4px+0⋅p²+y²=0

Which can also be written as −4px+y²=0

Now add 4px on both sides then the equation will become y²=4px.

The standard form for a horizontal parabola is y²=4px.

Page 176 Problem 9 Answer

It is required to explain why directrix places on the line x=−p.

Loci of all the points that are equidistant from a point called focus and a line called the directrix.

since the vertex of this parabola lines on the origin with its focus on (p,0); the directrix must be p units on the other sides of the origin and other side of the original and it is perpendicular to the axis of symmetry therefore x=−p is a directrix.

The directrix must be p units on the other sides of the origin and other side of the original and it is perpendicular to the axis of symmetry therefore x=−p is a directrix.

Page 176 Problem 10 Answer

It is required to explain why directrix places on the line x=−p.

The general equation of a parabola with its vertex on the origin is y²=4px. This is when the vertex is (0,0).

If the vertex lies on the point (h,k) which implies that (0,0) is shifted to (h,k).

Hence y is replaced by y−k and x with x−h.

Then the new equation will become (y−k)²=4p(x−h).

If the vertex lies on the point (h,k) which implies that (0,0) is shifted to (h,k).

Hence y is replaced by y−k and x with x−h.

Page 176 Problem 11 Answer

It is given the focus of parabola is located at (0,−2) and the equation of its directrix is y=2.

It is required to determine the equation of the parabola with vertex (0,0) and plot the parabola, the focus and the directrix.

This can be done by finding the general equation of the parabola then plot the graph accordingly.

We have to substitute values of h=0,k=0 and p=−2 in the equation (x−h)²=4p(y−k).

(x−0)²=4(−2)(y−0)

x²=−8y

Now plot a graph of the parabola, the focus and the directrix.

The equation of the parabola is x²=−8y.

It can be shown by a graph drawn below:

Page 177 Problem 12 Answer

It is given a parabola with focus(2,0) and directrix x=−2.

It is required to find the equation of parabola and plot it on graph.

To solve this check that the parabola is vertical or horizontal with the help of directrix.

After that use the suitable standard equation of parabola and place the value of a in standard equation to get the equation of parabola.

Given parabola have focus(2,0) and directrix x=−2. Since the directrix is parallel to y−axis so the parabola is horizontal parabola.

Now, standard equation of horizontal parabola is y²=4ax where (a,0) is the focus of parabola. Substituting the value of a in the equation,

y²=4⋅2⋅x

y²=8x​

Thus, equation of the parabola is,y²=8x

Plotting the parabola,

Therefore, equation of the parabola is,y²=8x

Graph of the parabola,

Page 177 Problem 13 Answer

It is given a parabola with focus(0,−1/2) and directrix y=1/2.

It is required to find the equation of parabola and plot it on graph.

To solve this check that the parabola is vertical or horizontal with the help of directrix.

After that use the suitable standard equation of parabola and place the value of a in standard equation to get the equation of parabola.

Given parabola have focus(0,−1/2) and directrix y=1/2.

Since the directrix is parallel to x−axis so, the parabola is vertical parabola.

Now, standard equation of horizontal parabola is x²=4ay where (0,a) is the focus of parabola.

Substituting the value of a in the equation,​x²=4⋅(−1/2)⋅y

x²=−2y​

Thus, equation of the parabola is,x²=−2y

Plotting the parabola,

Therefore, equation of the parabola is,x²=−2y Plotting the parabola,

Page 178 Problem 14 Answer

It is given the focus of parabola is located at (−1,−1) and the equation of its directrix is x=5.

It is required to determine the equation of the parabola with vertex (0,0)

and plot the parabola, the focus and the directrix.

This can be done by finding the general equation of the parabola then plot the graph accordingly.

It is known that, p=(x​value​of​focus)−(x​value​of​directrix)/2

p=−1−(5)/2

p=−3

Here, k= the y-coordinate of the focus   =−1.

The x−​value of the focus is h+p, so h+p=−1

h=−1+3

h=2

Substitute values of h=2,k=−1 and p=−3 in the equation(y−k)²=4p(x−h).

(y+1)²=4(−3)(x−2)(y+1)²

=−12(x−2)

Now plot a graph of the parabola, the focus and the directrix.

The equation of the parabola is (y+1)²=−12(x−2) and it can be drawn as:

Page 179 Problem 15 Answer

It is given a parabola with focus(5,−1) and directrix x=−3.

It is required to find the equation of parabola and plot it on graph.

To solve this check that the parabola is vertical or horizontal with the help of directrix.

After that use the suitable standard equation of parabola and place the value of a in standard equation to get the equation of parabola.

Given parabola have focus (5,−1) and directrix x=−3.

Since the directrix is parallel to y−axis so the parabola is horizontal parabola.

So, the standard equation of parabola is (y−k)²=4a(x−h).

For, value of a

a=5−(−3)/2

a=5+3/2

a=4

Now, for finding the value of (h,k)

k will be same as that of value of y−coordinate as focus have,k=−1

For the value of h

h+4=5

h=1

Now, substituting the value of (h,k) in equation.

Thus, equation of parabola is (y−(−1))²

=4⋅4⋅(x−1)(y+1)²

=16(x−1)​

Thus, equation of the parabola is,(y+1⁾²

=16(x−1)

Plotting the parabola,

Equation of the parabola is,(y+1)²=16(x−1)

Graph of the parabola is,

Page 179 Problem 16 Answer

It is given a parabola with focus(5,−1) and directrix x=−3.

It is required to find the equation of parabola and plot it on graph.

To solve this check that the parabola is vertical or horizontal with the help of directrix.

After that use the suitable standard equation of parabola and place the value of a in standard equation to get the equation of parabola.

Given parabola have focus(−2,0) and directrix y=4.

Since the directrix is parallel to x−axis so the parabola is vertical parabola.

So, the standard equation of parabola is (x−h)²

=4a(y−k)

So, focus will be (h,k+p)=(−2,0)

Thus, ​h=−2

k+a=0

Also, directrix

​y=k−a

4=k−a

So, the values

​k=2

a=−2

Now, the equation of parabola is,

(x−(−2))²

=4⋅(−2)(y−2)(x+2)²

=(−8)(y−2)​

Equation of the parabola is,(x+2)²

=(−8)(y−2)

Plotting the equation,

Therefore, equation of the parabola is,(x+2)²=(−8)(y−2)

Graph of the equation,

Page 179 Problem 17 Answer

It is given an equation y²+2x+8y+18=0.

It is required to convert the given equation into the standard form of a parabola and graph the parabola, the focus and the directrix.

This can be done by converting the given equation into its standard form then find the focus and equation of the directrix then plot the graph of required parabola.

Isolate the y terms.y²+8y=−2x−18

Add 4²  on both sides.

y²+8y+4²=−2x−18+16

Now convert the above equation into its standard form.​

(y+4)²=−2x−2(y+4)²=−2(x+1)​

Now on comparing the above equation with the standard equation (y−k)²=4p(x−h).

4p=−2

⇒p=−1/2

h=−1,

k=−4

Focus =(h+p,k)

=(−3/2,−4).

Directrix: x=h−p

=−1/2.

Based on the data above the graph must be drawn like:

The equation of the parabola is (y+4)²=−2(x+1) and this can be drawn as:

Page 191 Problem 18 Answer

It is given equation of parabola y²−12x−4y+64=0

It is required to convert the given equation in standard form of the parabola and graph the parabola.

To solve this, use complete square method then by further solving convert it in standard form.

Given equation of parabola is y²−12x−4y+64=0

y²−4y=12x−64

y²−2⋅2⋅y+4=12x−64+4

(y−2)²=12(x−5)

(y−2)²=4⋅3⋅(x−5)​

Equation of parabola is(y−2)²=12(x−5)

Graph of the parabola is,

Equation of parabola is (y−2)²=12(x−5)

Graph of the parabola is,

Page 181 Problem 19 Answer

It is given equation of parabola x²+8x−16y−48=0

It is required to convert the given equation in standard form of the parabola and graph the parabola.

To solve this, use complete square method then by further solving convert it in standard form.

Given equation of parabola is, x²+8x−16y−48=0

To solve this rearrange it the use complete square method the convert

x²+8x=16y−48

x²+2⋅4⋅x+16=16y−48+16

(x+4)²=16(x−2)(x+4)²

=4⋅4⋅(x−2)​

Equation of parabola is (x+4)²

=16(x−2)

Graph of the parabola is,

Equation of parabola is(x+4)²=16(x−2)

Graph of the parabola is,

Page 181 Problem 20 Answer

It is given that the focal length of a telescope is 140 mm and the mirror has a 70 mm diameter.

It is required to determine the depth of the bowl of the mirror.

This can be done by finding the equation of the parabola then substituting the value of y

with the point that lies on the telescope.

The distance from the bottom of the mirror’s bowl to the focus is p.

The vertex location is not specified, so use (0,0) for simplicity.

The equation for the mirror is a horizontal parabola (with x the distance along the telescope and y the position out from the centre).

(y−0)²=4p(x−0)

Now p is equal to 140 mm as it is the focal length.

y²=540x

Since the diameter of the bowl of the mirror is 70 mm, the points at the rim of the mirror have why values of 35 mm and −35 mm.

The x- value of either point will be the same as the x-value of the point directly above the bottom of the bowl, which equals the depth of the bowl.

Since the points on the rim lie on the parabola use the equation of the parabola to solve for the x-value of either edge of the mirror.

352

=560x

⇒x=2.1875mm​

The bowl is approximately 2.19 mm deep.

Page 182 Problem 21 Answer

A football team needs one more field goal to win the game. The goalpost that the ball must clear is 10

feet off the ground. The path of the football after kicked for a 35−yard goal is given by the equation (y−11)=−0.0125(x−20)²

It is given equation and the point (35,10/3).

The horizontal distance to covered by the ball is 35 yard and the vertical distance must be at least 10 feet or 10/3 yard to pass the goal post therefore

It is required to find the equation of parabola and plot it on graph.

The trajectory of the ball is such that is passes above the goalpost which means that the team scored the field goal and therefore won the game.

Page 183 Problem 22 Answer

It is given the focus and the vertex of the parabola.

It is required to determine a relationship between the separation of the focus, vertex, and the shape of the parabola.

If the focus is located at the y-axis, then the parabola must be vertical and vice versa.

If the focus is located at the x-axis, then the parabola must be horizontal and vice versa.

In the equation x²=4py, y=ax²

is only possible if a=1/4p.

There must be a vertical parabola when the focus is located at the y-axis while the parabola must be horizontal when it’s focus is located at the x-axis.

Page 183 Problem 23 Answer

It is given two variables x and y.

It is required to derive an equation relating x and y from the definition of a parabola based on focus and directrix by using the distance formula.

This can be done by finding the distance between the point on the parabola and the focus.

Then calculate the distance between the point on the parabola and the directrix then equating both distances to derive the equation of the parabola.

Let a parabola that open upwards or downwards, then the directrix will be a horizontal line of the form y=c.

Let (a,b) be the focus and let y=c be the directrix.

Let (x₀,y₀) be any point on the parabola.

Now calculate the distance between the point (x₀,y₀) and (a,b).

√(x₀−a)²+(y₀−b)²

Now calculate the distance between the point (x₀,y₀) and the line y=c.

∣y₀−c∣

Here, the distance between the point and the line is the difference between their y-coordinates.

Now equate the above two equations.

√(x₀−a)²+(y₀−b)²

=∣y₀−c∣

Square both sides.

(x₀−a)²+(y₀−b)²=(y₀−c)²

Now simplify the equation.

(x₀−a)²+b²−c²=2(b−c)y₀

The equation in (x₀,y₀) is true for all other values on the parabola and hence on the parabola And hence rewrite the equation with (x,y).

Therefore, the equation will be: (x−a)²+b²−c²=2(b−c)y

The equation of the parabola is (x−a)²+b²−c²=2(b−c)y.

Page 183 Exercise 1 Answer

It is given the focus of parabola is located at (3,0) and the equation of its directrix is x=−3.

It is required to determine the equation of the parabola with vertex (0,0) and plot the parabola, the focus and the directrix.

This can be done by finding the general equation of the parabola then plot the graph accordingly.

Substitute values of ​h=0,k=0 and p=3 in the equation

(y−k)²=4p(x−h).

(y−0)²=4(3)(x−0)

y²⁼12x

Now plot a graph of the parabola, the focus and the directrix.

The equation of the parabola is y²=12x.

Page 183 Exercise 2 Answer

It is given the focus of parabola is located at (0,2) and the equation of its directrix is y=−2.

It is required to determine the equation of the parabola with vertex (0,0) and plot the parabola, the focus and the directrix.

This can be done by finding the general equation of the parabola then plot the graph accordingly.

We have to substitute values of h=0,k=0 and p=2 in the equation

(x−h)²=4p(y−k).

(x−0)²=4(2)(y−0)x²=8y​

Now plot a graph of the parabola, the focus and the directrix.

The equation of the parabola is x²=8y.

Page 184 Exercise 3 Answer

It is given that the vertex of a parabola is at (−3,6) and directrix is x=−1.75.

It is required to find the equation of the parabola.

The equation of the directrix of a parabola is given as x=−p.

To find the equation of the parabola, substitute (−3,6) for (h,k), 1.75

for p in the equation (y−k)²=4p(x−h).

We have to substitute (−3,6) for (h,k), 1.75

for p in the equation (y−k)²=4p(x−h).

Simplify the equation by using the identity (a+b)²=a²+2ab+b².

​​(y−6)²=4(1.75)(x−(−3))

y²−12y+36=(7)(x+3)

y²−12y+36=7x+21

y²−12y−7x+15=0

​The equation of the parabola with vertex (−3,6) and the given directrix is y²−12y−7x+15=0.

Page 184 Exercise 4 Answer

It is given that the vertex of a parabola is at (6,20) and focus at (6,11).

It is required to find the equation of the parabola.

To find the equation of the parabola, substitute (6,20)

for (h,k), (6,11)

for (h,p) in the equation (x−h)²

=4(p−k)(y−k).

We have to substitute (6,20)

for (h,k), (6,11)

for (h,p) in the equation (x−h)²

=4(p−k)(y−k).

Simplify the equation by using the identity (a+b)²=a²+2ab+b².

(x−6)²=4(11−20)(y−20)

x²−12x+36=4(−9)(y−20)

x²−12x+36=−36y+720

x²−12x+36y−684=0​

The equation of the parabola with vertex (6,20) and focus at (6,11) is x²−12x+36y−684=0.

Page 184 Exercise 5 Answer

It is given that the focus of a parabola is at (5,3) and directrix is x=7.

It is required to find the equation of the parabola.

To find the equation of the parabola, substitute (5,3)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)2+(y₂−y₁)².

Then, substitute (x,7)

for (x1,y1), (x,y)

for (x2,y2) in the formula d=√(x₂−x₁)²+(y²−y₁)².

Equate the two distances and simplify the equation further.

We have to substitute (5,3)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)²+(y₂−y₁)².

d=√(x−5)²+(y−3)²

d=√x²−10x+25+y²−6y+9

We have to substitute (x,7)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)²

+(y₂−y₁)².​

d=√(x−x)²+(y−7)²

d=√0+y²−14y+49

Equate the two distance. Simplify the equation by squaring on both sides. ​

√x²−10x+25+y²−6y+9

=√0+y²−14y+49

x²−10x+25+y²−6y+9=y²−14y+49x²−10x−8y−15=0​

We have to plot the equation x²−10x−8y−15=0 by using a graphing tool.

The equation of the parabola with focus (5,3) and the given directrix is x²−10x−8y−15=0.

The graph of the parabola is as follows:

Page 184 Exercise 6 Answer

It is given that the focus of a parabola is at (−3,3) and directrix is x=3.

It is required to find the equation of the parabola.

To find the equation of the parabola, substitute (−3,3)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)²+(y₂−y₁)².

Then, substitute (x,7)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)²+(y₂−y₁)².

Equate the two distances and simplify the equation further.

We have to substitute (−3,3)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)²+(y₂−y₁)².​

d=√(x−(−3))²+(y−3)²

d=√x²+6x+9+y²−6y+9

We have to substitute (3,y)

for (x₁,y₁), (x,y)

for (x₂,y₂) in the formula d=√(x₂−x₁)²+(y₂−y₁)².​

d=√(x−3)²+(y−y)²

d=√x²−6x+9

Equate the two distance. Simplify the equation by squaring on both sides.​

√x²+6x+9+y²−6y+9

=√x²−6x+9

x²+6x+9+y²−6y+9=x²−6x+9y²−6x+12x+9=0

Plot the equation y²−6x+12x+9=0 by using a graphing tool.

The equation of the parabola with focus (5,3) and the given directrix is y²−6x+12x+9=0

The graph of the parabola is as follows:

 

Page 184 Exercise 7 Answer

The equation of a parabola is given as y²−20x−6y−51=0.

It is required to write the equation in standard form and the plot the parabola along with its focus and directrix.

Write the given equation in the form (y−k)²=4p(x−h).

Then, the vertex of the parabola is (h,k), its focus is (h+p,k) and the equation of the directrix isx=h−p.

The equation can be written as (y²−6y+9)−20(x+51/20)=9.

On simplifying the equation,

(y²−6y+9)=20(x+51/20+9/20)

(y−3)²=4(5)(x−(−3))

This means that the vertex of the parabola is at (−3,3), the focus is at (2,3) and the equation of the directrix is x=−8.

We have to plot the equation (y−3)²=4(5)(x−(−3)) by using a graphing tool.

The standard form of the given equation (y−3)²=4(5)(x−(−3)).

The plot of the parabola is:

Page 184 Exercise 8 Answer

The equation of a parabola is given as x²−14x−12y+73=0.

It is required to write the equation in standard form and the plot the parabola along with its focus and directrix.

Write the given equation in the form (x−h)²=4p(y−k).

Then, the vertex of the parabola is (h,k), its focus is (h,k+p) and the equation of the directrix is y=k−p.

The equation can be written as (x²−14x+49)−12(y−73/12)=49.

On simplifying the equation,

(x²−14x+49)=12(y−73/12+49/12)

(x−7)²=12(y−24/12)

(x−7)²=4(3)(y−2)​

This means that the vertex of the parabola is at (7,2), the focus is at (7,5) and the equation of the directrix is y=−1.

We have to plot the equation (x−7)²=4(3)(y−2) by using a graphing tool.

The standard form of the given equation (x−7)²=4(3)(y−2).

The plot of the parabola is:

Page 184 Exercise 9 Answer

It is given that the equation of the cross-section of a parabolic satellite dish is y=1/50x².

It is required to find the distance between the focus and the vertex of the cross-section.

If the equation of the parabola is of the form x²=4py, then the distance between the focus and the vertex is p.

Here, the given equation is x²=4(12.5)y.

Therefore, the distance between the focus and the vertex is 12.5 inches.

The focus of the satellite dish is 12.5 inches far from the vertex of the cross-section.

Page 185 Exercise 10 Answer

It is a given equation of parabola y+1=1/16(x−2)².

It is required to convert the given equation in standard form of the parabola and graph the parabola.

To solve this, convert it in standard form. Then plot the correct graph for the given equation.

It is a given equation of parabola y+1=1/16(x−2)².

Now to convert equation in standard form multiply 16,

Then, y+1=1/16(x−2)²

16(y+1)=(x−2)²

4⋅4(y+1)=(x−2)²

So, the equation is,(x−2)²=16(y+1)

By comparing the obtained equation with standard equation,​

h=2 k=−1​

Now plotting the parabola on graph,

The equation will be (x−2)²=16(y+1).

Graph of the parabola is,

Page 185 Exercise 11 Answer

It is given equation of parabola y−1=1/16(x+2)²

It is required to convert the given equation in standard form of the parabola and graph the parabola.

To solve this, convert it in standard form. Then plot the correct graph for the given equation.

It is given equation of parabola y−1=1/16(x+2)²

Converting equation in standard form,

y−1=1/16(x+2)²

16(y−1)=(x+2)²

4⋅4(y−1)=(x+2)²

So, the equation is, (x+2)²=4⋅4⋅(y−1)

By comparing the obtained equation with standard equation,​

h=−2 k=1​

Now plotting the parabola on graph,

The equation is, (x+2)²=4⋅4⋅(y−1)

Graph of the parabola is,

Page 185 Exercise 12 Answer

It is a given equation of parabola x+1=−1/16(y−2)².

It is required to convert the given equation in standard form of the parabola and graph the parabola.

To solve this, convert it in standard form. Then plot the correct graph for the given equation.

The given equation of parabola is x+1=−1/16(y−2)².

Converting equation in standard form,​

x+1=−1/16(y−2)² 

−16(x+1)=(y−2)²

−4⋅4(x+1)=(y−2)²

So, the equation is,(y−2)²=−16(x+1)

By comparing the obtained equation with standard equation,​

h=−1 k=2​

Now plotting the parabola on graph,

The equation is,(y−2)²=−16(x+1)

Graph of the parabola is,

Page 186 Exercise 13 Answer

It is given an upward- opening parabola with focus(−p,0) and directrix x=p.

It is required to find the equation of the parabola.

To solve this, find the vertex of parabola then substitute it on the standard equation of parabola to get the equation of parabola.

Given parabola has focus(−p,0) and directrix x=p.

From this vertex will be,(h,k)=(0,0)

So, h=0 k=0

Now, substituting the values on standard equation,

(y−0)²=4⋅p⋅(x−0)

y²=4px​

Equation of the parabola is, y²=4px.

Page 186 Exercise 14 Answer

It is given that a ball is projected in a parabolic path y−4=−4/1521(x−39)².

The tennis net is $3feet$ high and the total length of the court is 78 feet.

The position of player is x=0

It is required to find out how far the net is located from the player.

To solve this, find the distance between the player and net which will be equal to half of the total length of the court.

The tennis net is 3 feet high and the total length of the court is 78 feet.

The position of player is x=0

Now, the distance between the players and net is,1/2×78=39 feet

So, distance between the player and net =39 feet.

Page 186 Exercise 15 Answer

It is given that a ball is projected in a parabolic path y−4=−4/1521(x−39)².

The tennis net is 3 feet high and the total length of the court is 78 feet.

The position of player is x=0

It is required to show why the ball will go over the net.

To solve this, write the equation in standard form then find the value of vertical distance y by substituting the value of x=39.

Since, the net is at x=39.

Given equation is,​ y−4=−4/1521(x−39)²

(x−39)²=−1521/4(y−4)

Substituting the value x=39 to get the value of y,

(39−39)²=−1521/4

(y−4)0=−1521/4(y−4)

y−4=0

y=4

Now the height of the net is 3 feet but the ball goes to a height of 4 feet.

Ball does not touch the net because the height of the net is 3 feet but the ball goes to a height of 4 feet.

Page 186 Exercise 16 Answer

It is given that a ball is projected in a parabolic path y−4=−4/1521(x−39)².

The tennis net is 3 feet high and the total length of the court is 78 feet.

The position of player is x=0

It is required to show that the ball lands either inside the court or on the opposite end line.

To solve this, find the range of the trajectory of the ball.

Given equation is, y−4=−4/1521(x−39)²

(x−39)²=−1521/4(y−4)

Total range of the ball is, 2⋅39=78 feet

So, the ball will land on the opposite end line on the other side of the court.

The ball will land on the opposite end line on the other side of the court.

Page 187 Exercise 17 Answer

It is required to find the length of the latus rectum of the parabola.

To solve this let’s consider a parabola y²=4ax which has vertex (0,0) and focus (a,0).

Now find the endpoints of the latus rectum which will lie vertically above and below the focus.

So, there x− coordinate will be a.

Then substitute a in place of x to get the value of y for the points. Then find the distance between the points.

Let’s consider a parabola y²=4ax which has vertex (0,0) and focus (a,0).

Then x− coordinate of the latus rectum is a. Now solving for y− coordinate​

y²=4ax

y²=4⋅a⋅a

y²=(2a)²

y=±2a

So, the endpoints of latus rectum are (a,2a),(a,−2a)

Since x− coordinate of the points are the same, taking the distance between the endpoints of the latus rectum is 4a by laterally solving.

So, length of latus rectum =4a

Length of latus rectum =4a