# Eureka Math Precalculus Module 3 Lesson 8 Answer Key

## Engage NY Eureka Math Precalculus Module 3 Lesson 8 Answer Key

### Eureka Math Precalculus Module 3 Lesson 8 Exercise Answer Key

Exercises
Exercise 1.
Let F(0, 5) and G(0, – 5) be the foci of a hyperbola. Let the points P(x,y) on the hyperbola satisfy either PF – PG = 6 or PG – PF = 6. Use the distance formula to derive an equation for this hyperbola, writing your answer in the form $$\frac{x^{2}}{a^{2}}$$ – $$\frac{y^{2}}{b^{2}}$$ = 1.
PG – PF = 6
$$\sqrt{x^{2} + (y + 5)^{2}}$$ – $$\sqrt{x^{2} + (y – 5)^{2}}$$ = 6
$$\sqrt{x^{2} + (y + 5)^{2}}$$ = 6 + $$\sqrt{x^{2} + (y – 5)^{2}}$$
x2 + (y2 + 10y + 25) = 36 + 12$$\sqrt{x^{2} + (y – 5)^{2}}$$ + (x2 + y2 – 10y + 25)
20y – 36 = 12$$\sqrt{x^{2} + (y – 5)^{2}}$$
5y – 9 = 3$$\sqrt{x^{2} + (y – 5)^{2}}$$
25y2 – 90y + 81 = 9(x2 + y2 – 10y + 25)
25y2 – 90y + 81 = 9x2 + 9y2 – 90y + 225
16y2 – 9x2 = 144
$$\frac{y^{2}}{9}$$ – $$\frac{x^{2}}{16}$$ = 1

Exercise 2.
Where does the hyperbola described above intersect the y – axis?
The curve intersects the y – axis at (0,3) and (0, – 3).

Exercise 3.
Find an equation for the line that acts as a boundary for the portion of the curve that lies in the first quadrant.
For large values of x, $$\frac{y}{3}$$≈$$\frac{x}{4}$$, so the line y = $$\frac{3}{4}$$ x is the boundary for the curve in the first quadrant.

Exercise 4.
Sketch the graph of the hyperbola described above. ### Eureka Math Precalculus Module 3 Lesson 8 Problem Set Answer Key

Question 1.
For each hyperbola described below: (1) Derive an equation of the form $$\frac{x^{2}}{a^{2}}$$ – $$\frac{y^{2}}{b^{2}}$$ = 1 or $$\frac{y^{2}}{b^{2}}$$ – $$\frac{x^{2}}{a^{2}}$$ = 1. (2) State any x – or y – intercepts. (3) Find the equations for the asymptotes of the hyperbola.
a. Let the foci be A( – 2,0) and B(2,0), and let P be a point for which either PA – PB = 2 or PB – PA = 2.
i. x2 – $$\frac{y^{2}}{3}$$ = 1
ii. ( – 1,0), (1,0); no y – intercepts
iii. y≈$$\sqrt{3}$$ x, so y = ±$$\sqrt{3}$$ x

b. Let the foci be A( – 5,0) and B(5,0), and let P be a point for which either PA – PB = 5 or PB – PA = 5.
i. $$\frac{x^{2}}{\frac{25}{4}}$$ – $$\frac{y^{2}}{\frac{75}{4}}$$ = 1
ii. ( – 2.5,0), (2.5,0); no y – intercepts
iii. y≈$$\frac{\sqrt{75}}{2}$$⋅$$\frac{2}{5}$$ x = $$\sqrt{3}$$ x, so y = ±$$\sqrt{3}$$ x

c. Consider A(0, – 3) and B(0,3), and let P be a point for which either PA – PB = 2.5 or PB – PA = 2.5.
i. $$\frac{y^{2}}{\frac{9}{4}}$$ – $$\frac{x^{2}}{\frac{27}{4}}$$ = 1
ii. (0,1.5), (0, – 1.5); no x – intercepts
iii. y≈$$\frac{3}{2}$$⋅$$\frac{2}{\sqrt{27}}$$ x = $$\frac{1}{\sqrt{3}}$$x, so y = ±$$\frac{\sqrt{3}}{3}$$x

d. Consider A(0, – $$\sqrt{2}$$) and B(0,$$\sqrt{2}$$), and let P be a point for which either PA – PB = 2 or PB – PA = 2.
i. y2 – x2 = 1
ii. (0,1), (0, – 1); no x – intercepts.
iii. y = ±x

Question 2.
Graph the hyperbolas in parts (a)–(d) in Problem 1.
a. b. c. d. Question 3.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
x2 – y2 = k.
a. k = 4 b. k = 1 c. k = $$\frac{1}{4}$$ d. k = 0 e. k = – $$\frac{1}{4}$$ f. k = – 1 g. k = – 4 h. Describe the hyperbolas x2 – y2 = k for different values of k. Consider both positive and negative values of k, and consider values of k close to zero and far from zero.
If k is close to zero, then the hyperbola is very close to the asymptotes y = x and y = – x, appearing almost to have corners as the graph crosses the x – axis. If k is far from zero, the hyperbola gets farther from the asymptotes near the center. If k > 0, then the hyperbola crosses the x – axis, opening to the right and left, and if k<0, then the hyperbola crosses the y – axis, opening up and down.

i. Are there any values of k so that the equation x2 – y2 = k has no solution?
No. The equation x2 – y2 = k always has solutions. The solution points lie on either two intersecting lines or on a hyperbola.

Question 4.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
$$\frac{x^{2}}{k}$$ – y2 = 1.
a. k = – 1
There is no solution to the equation $$\frac{x^{2}}{ – 1}$$ – y2 = 1, because there are no real numbers x and y so that x2 + y2 = – 1.

b. k = 1 c. k = 2 d. k = 4 e. k = 10 f. k = 25 g. Describe what happens to the graph of $$\frac{x^{2}}{k}$$ – y2 = 1 as k → ∞.
As k → ∞, it appears that the hyperbolas with equation $$\frac{x^{2}}{k}$$ – y2 = 1 get flatter; the x – intercepts get farther from the center at the origin, and the asymptotes get less steep.

Question 5.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
x2 – $$\frac{y^{2}}{k}$$ = 1.
a. k = – 1 b. k = 1 c. k = 2 d. k = 4 e. k = 10 f. Describe what happens to the graph x2 – $$\frac{y^{2}}{k}$$ = 1 as k → ∞.
As k → ∞, the hyperbola with equation x2 – $$\frac{y^{2}}{k}$$ = 1 is increasingly stretched vertically. The center and intercepts do not change, but the steepness of the asymptotes increases.

Question 6.
An equation of the form ax2 + bx + cy2 + dy + e = 0 where a and c have opposite signs might represent a hyperbola.
a. Apply the process of completing the square in both x and y to convert the equation
9x2 – 36x – 4y2 – 8y – 4 = 0 to one of the standard forms for a hyperbola: $$\frac{(x – h)^{2}}{a^{2}}$$ – $$\frac{(y – k)^{2}}{b^{2}}$$ = 1 or $$\frac{(y – k)^{2}}{b^{2}}$$ – $$\frac{(x – h)^{2}}{a^{2}}$$ = 1.
9x2 – 36x – 4y2 – 8y = 4
9(x2 – 4x) – 4(y2 + 2y) = 4
9(x2 – 4x + 4) – 4(y2 + 2y + 1) = 4 + 36 – 4
9(x – 2)2 – 4(y + 1)2 = 36
$$\frac{(x – 2)^{2}}{4}$$ – $$\frac{(y + 1)^{2}}{9}$$ = 1

b. Find the center of this hyperbola.
The center is (2, – 1).

c. Find the asymptotes of this hyperbola.
$$\frac{x – 2}{2}$$ = ±$$\frac{y + 1}{3}$$
y = $$\frac{3}{2}$$ x – 4 or y = – $$\frac{3}{2}$$ x + 2

d. Graph the hyperbola. Question 7.
For each equation below, identify the graph as either an ellipse, a hyperbola, two lines, or a single point. If possible, write the equation in the standard form for either an ellipse or a hyperbola.
a. 4x2 – 8x + 25y2 – 100y + 4 = 0
In standard form, this is the equation of an ellipse: $$\frac{(x – 1)^{2}}{25}$$ + $$\frac{(y – 2)^{2}}{4}$$ = 1.

b. 4x2 – 16x – 9y2 – 54y – 65 = 0
When we try to put this equation in standard form, we find $$\frac{(x – 2)^{2}}{9}$$ – $$\frac{(y + 3)^{2}}{4}$$ = 0, which gives
($$\frac{x – 2}{3}$$ = ±$$\frac{y + 3}{2}$$. These are the lines with equation y = $$\frac{1}{3}$$ (2x – 13) and y = $$\frac{1}{3}$$ ( – 2x – 5).

c. 4x2 + 8x + y2 + 2y + 5 = 0
When we try to put this equation in standard form, we find (x + 1)2 + $$\frac{(y + 1)^{2}}{4}$$ = 0. The graph of this equation is the single point ( – 1, – 1).

d. – 49x2 + 98x + 4y2 – 245 = 0
In standard form, this is the equation of a hyperbola: $$\frac{y^{2}}{49}$$ – $$\frac{(x – 1)^{2}}{4}$$ = 1.

e. What can you tell about a graph of an equation of the form ax2 + bx + cy2 + dy + e = 0 by looking at the coefficients?
There are two categories; if the coefficients a and c have the same sign, then the graph is either an ellipse, a point, or an empty set. If the coefficients a and c have opposite signs, then the graph is a hyperbola or two intersecting lines. We cannot tell just by looking at the coefficients which of these sub – cases hold.

### Eureka Math Precalculus Module 3 Lesson 8 Exit Ticket Answer Key

Question 1.
Let F( – 4,0) and B(4,0) be the foci of a hyperbola. Let the points P(x,y) on the hyperbola satisfy either PF – PG = 4 or PG – PF = 4. Derive an equation for this hyperbola, writing your answer in the form $$\frac{x^{2}}{a^{2}}$$ – $$\frac{y^{2}}{b^{2}}$$ = 1.
PF = $$\sqrt{(x + 4)^{2} + y^{2}}$$
PB = $$\sqrt{(x – 4)^{2} + y^{2}}$$
$$\sqrt{(x + 4)^{2} + y^{2}}$$ = 4 + $$\sqrt{(x – 4)^{2} + y^{2}}$$
(x + 4)2 + y2 = 16 + 8$$\sqrt{(x – 4)^{2} + y^{2}}$$ + (x – 4)2 + y2
x2 + 8x + 16 + y2 = 16 + 8$$\sqrt{(x – 4)^{2} + y^{2}}$$ + x2 – 8x + 16 + y2
16x – 16 = 8$$\sqrt{(x – 4)^{2} + y^{2}}$$
2x – 2 = $$\sqrt{(x – 4)^{2} + y^{2}}$$
$$\frac{x^{2}}{4}$$ – $$\frac{y^{2}}{12}$$ = 1