# Eureka Math Precalculus Module 3 Lesson 7 Answer Key

## Engage NY Eureka Math Precalculus Module 3 Lesson 7 Answer Key

### Eureka Math Precalculus Module 3 Lesson 7 Exercise Answer Key

Exercise
Points F and G are located at (0, 3) and (0, – 3). Let P(x, y) be a point such that PF + PG = 8. Use this information to show that the equation of the ellipse is $$\frac{x^{2}}{7} + \frac{y^{2}}{16}$$ = 1. The distance from the x – axis to point F and to point G is 3. The distance from the x – axis to point P is x; that means the vertical distance from F to P is 3 – y, and the vertical distance from G to P is 3 + y.
PF + PG = $$\sqrt{x^{2} + (3 – y)^{2}}$$ + $$\sqrt{x^{2} + (y + 3)^{2}}$$ = 8
$$\sqrt{x^{2} + (3 – y)^{2}}$$ = 8 – $$\sqrt{x^{2} + (y + 3)^{2}}$$
x2 + (3 – y)2 = 64 – 16$$\sqrt{x^{2} + (y + 3)^{2}}$$ + x2 + (y + 3)2
y2 – 6y + 9 = 64 – 16$$\sqrt{x^{2} + (y + 3)^{2}}$$ + y2 + 6y + 9
16$$\sqrt{x^{2} + (y + 3)^{2}}$$ = 64 + 12y
256[x2 + (y + 3)2 ] = 4096 + 1536y + 144y2
256[x2 + y2 + 6y + 9] = 4096 + 1536y + 144y2
256x2 + 256y2 + 1536y + 2304 = 4096 + 1536y + 144y2
256x2 + 112y2 = 1792
$$\frac{x^{2}}{7} + \frac{y^{2}}{16}$$ = 1

### Eureka Math Precalculus Module 3 Lesson 7 Problem Set Answer Key

Question 1.
Derive the equation of the ellipse with the given foci F and G that passes through point P. Write your answer in standard form:  = 1.
a. The foci are F( – 2, 0) and G(2, 0), and point P(x, y) satisfies the condition PF + PG = 5.
$$\frac{x^{2}}{\frac{25}{4}}$$ + $$\frac{y^{2}}{\frac{9}{4}}$$ = 1
$$\frac{4 x^{2}}{25}$$ + $$\frac{4 y^{2}}{9}$$

b. The foci are F( – 1, 0) and G(1, 0), and point P(x, y) satisfies the condition PF + PG = 5.
$$\frac{x^{2}}{\frac{25}{4}}$$ + $$\frac{y^{2}}{\frac{21}{4}}$$ = 1
$$\frac{4 x^{2}}{25}$$ + $$\frac{4 y^{2}}{21}$$

c. The foci are F(0, – 1) and G(0, 1), and point P(x, y) satisfies the condition PF + PG = 4.
$$\frac{x^{2}}{3}$$ + $$\frac{y^{2}}{4}$$ = 1

d. The foci are F( – $$\frac{2}{3}$$, 0) and G($$\frac{2}{3}$$, 0), and point P(x, y) satisfies the condition PF + PG = 3.
$$\frac{x^{2}}{\frac{9}{4}}$$ + $$\frac{y^{2}}{\frac{65}{36}}$$ = 1
$$\frac{4x^{2}}{9}$$ + $$\frac{36y^{2}}{65}$$ = 1

e. The foci are F(0, – 5) and G(0, 5), and point P(x, y) satisfies the condition PF + PG = 12.
$$\frac{x^{2}}{11}$$ + $$\frac{y^{2}}{36}$$ = 1

f. The foci are F( – 6, 0) and G(6, 0), and point P(x, y) satisfies the condition PF + PG = 20.
$$\frac{x^{2}}{100}$$ + $$\frac{y^{2}}{64}$$ = 1

Question 2.
Recall from Lesson 6 that the semi – major axes of an ellipse are the segments from the center to the farthest vertices, and the semi – minor axes are the segments from the center to the closest vertices. For each of the ellipses in Problem 1, find the lengths a and b of the semi – major axes.
a. a = $$\frac{5}{2}$$, b = $$\frac{3}{2}$$
b. a = $$\frac{5}{2}$$, b = $$\frac{\sqrt{21}}{2}$$
c. a = $$\sqrt{3}$$, b = 2
d. a = $$\frac{3}{2}$$, b = $$\frac{\sqrt{65}}{6}$$
e. a = $$\sqrt{11}$$, b = 6
f. a = 10, b = 8

Question 3.
Summarize what you know about equations of ellipses centered at the origin with vertices (a, 0), ( – a, 0), (0, b), and (0, – b).
For ellipses centered at the origin, the equation is always $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1, where a is the positive x – value of the
x – intercepts and b is the positive y – value of the y – intercepts. If we know the x – and y – intercepts, then we know the equation of the ellipse.

Question 4.
Use your answer to Problem 3 to find the equation of the ellipse for each of the situations below.
a. An ellipse centered at the origin with x – intercepts ( – 2, 0), (2, 0) and y – intercepts (0, 8), (0, – 8)
$$\frac{x^{2}}{4}$$ + $$\frac{y^{2}}{64}$$ = 1

b. An ellipse centered at the origin with x – intercepts ( – $$\sqrt{5}$$, 0), ($$\sqrt{5}$$, 0) and y – intercepts (0, 3), (0, – 3)
$$\frac{x^{2}}{5}$$ + $$\frac{y^{2}}{9}$$ = 1

Question 5.
Examine the ellipses and the equations of the ellipses you have worked with, and describe the ellipses with equation $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 in the three cases a > b, a = b, and b > a.
If a > b, then the foci are on the x – axis and the ellipse is oriented horizontally, and if b > a, the foci are on the y – axis and the ellipse is oriented vertically. If a = b, then the ellipse is a circle with radius a.

Question 6.
Is it possible for $$\frac{x^{2}}{4}$$ + $$\frac{y^{2}}{9}$$ = 1 to have foci at ( – c, 0) and (c, 0) for some real number c?
No. Since 9 > 4, the foci must be along the y – axis.

Question 7.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
$$\frac{x^{2}}{4}$$ + y2 = k.
a. k = 1 b. k = $$\frac{1}{4}$$ c. k = $$\frac{1}{9}$$ d. k = $$\frac{1}{16}$$ e. k = $$\frac{1}{25}$$ f. k = $$\frac{1}{100}$$ g. Make a conjecture: Which points in the plane satisfy the equation $$\frac{x^{2}}{4}$$ + y2 = 0?
As k is getting smaller, the ellipse is shrinking. It seems that the only point that lies on the curve given by $$\frac{x^{2}}{4}$$ + y2 = 0 would be the single point (0, 0).

h. Explain why your conjecture in part (g) makes sense algebraically.
Both $$\frac{x^{2}}{4}$$ and y2 are nonnegative numbers, and the only way to sum two nonnegative numbers and get zero would be if they were both zero. Thus, $$\frac{x^{2}}{4}$$ = 0 and y2 = 0, which means that (x, y) is (0, 0).

i. Which points in the plane satisfy the equation $$\frac{x^{2}}{4}$$ + y2 = – 1?
There are no points in the plane that satisfy the equation x2 + y2 = – 1 because $$\frac{x^{2}}{4}$$ + y2 ≥ 0 for all real numbers x and y.

Question 8.
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 b. k = 2 c. k = 4 d. k = 10 e. k = 25 f. Describe what happens to the graph of $$\frac{x^{2}}{k}$$ + y2 = 1 as k→∞.
As k gets larger and larger, the ellipse stretches more and more horizontally, while not changing vertically. The vertices are ( – $$\sqrt{k}$$, 0), ($$\sqrt{k}$$, 0), (0, – 1), and (0, 1).

Question 9.
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 = 2 c. k = 4 d. k = 10 e. k = 25 f. Describe what happens to the graph of x2 + $$\frac{y^{2}}{k}$$ = 1 as k → ∞.
As k gets larger and larger, the ellipse stretches more and more vertically, while not changing horizontally. The vertices are ( – 1, 0), (1, 0), (0, – $$\sqrt{k}$$), and (0, $$\sqrt{k}$$).

### Eureka Math Precalculus Module 3 Lesson 7 Exit Ticket Answer Key

Question 1.
Suppose that the foci of an ellipse are F( – 1, 0) and G(1, 0) and that the point P(x, y) satisfies the condition
PF + PG = 4.
a. Derive an equation of an ellipse with foci F and G that passes through P. Write your answer in standard form: $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$$ = 1.
PF + PG = 4
$$\sqrt{(x – 1)^{2} + y^{2}}$$ + $$\sqrt{(x + 1)^{2} + y^{2}}$$ = 4
(x – 1)2 + y2 = 16 – 8$$\sqrt{(x + 1)^{2} + y^{2}}$$ + (x + 1)2 + y2
(x – 1)2 – (x + 1)2 – 16 = – 8$$\sqrt{(x + 1)^{2} + y^{2}}$$
x2 – 2x + 1 – (x2 + 2x + 1) – 16 = – 8$$\sqrt{(x + 1)^{2} + y^{2}}$$
– 4x – 16 = – 8$$\sqrt{(x + 1)^{2} + y^{2}}$$
x + 4 = 2$$\sqrt{(x + 1)^{2} + y^{2}}$$
x2 + 8x + 16 = 4(x2 + 2x + 1 + y2 )
x2 + 8x + 16 = 4x2 + 8x + 4 + 4y2
– 3x2 – 4y2 = – 12
$$\frac{x^{2}}{4} + \frac{y^{2}}{3}$$ = 1

b. Sketch the graph of the ellipse defined above.  For the y – intercepts (0, $$\sqrt{3}$$) and (0, – $$\sqrt{3}$$), we have $$\sqrt{\sqrt{3}^{2} + ( – 1)^{2}}$$ + $$\sqrt{\sqrt{3}^{2} + 1^{2}}$$ = 2 + 2 = 4 and $$\sqrt{( – \sqrt{3})^{2} + ( – 1)^{2}}$$ + $$\sqrt{( – \sqrt{3})^{2} + 1^{2}}$$ = 2 + 2 = 4.