## 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.

Answer:

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 – \(\sqrt{x^{2} + (y + 3)^{2}}\)

x^{2} + (3 – y)^{2} = 64 – 16\(\sqrt{x^{2} + (y + 3)^{2}}\) + x^{2} + (y + 3)^{2}

y^{2} – 6y + 9 = 64 – 16\(\sqrt{x^{2} + (y + 3)^{2}}\) + y^{2} + 6y + 9

16\(\sqrt{x^{2} + (y + 3)^{2}}\) = 64 + 12y

256[x^{2} + (y + 3)^{2} ] = 4096 + 1536y + 144y^{2}

256[x^{2} + y^{2} + 6y + 9] = 4096 + 1536y + 144y^{2}

256x^{2} + 256y^{2} + 1536y + 2304 = 4096 + 1536y + 144y^{2}

256x^{2} + 112y^{2} = 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.

Answer:

\(\frac{x^{2}}{\frac{25}{4}}\) + \(\frac{y^{2}}{\frac{9}{4}}\) = 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.

Answer:

\(\frac{x^{2}}{\frac{25}{4}}\) + \(\frac{y^{2}}{\frac{21}{4}}\) = 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.

Answer:

\(\frac{x^{2}}{3}\) + \(\frac{y^{2}}{4}\) = 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.

Answer:

\(\frac{x^{2}}{\frac{9}{4}}\) + \(\frac{y^{2}}{\frac{65}{36}}\) = 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.

Answer:

\(\frac{x^{2}}{11}\) + \(\frac{y^{2}}{36}\) = 1

f. The foci are F( – 6, 0) and G(6, 0), and point P(x, y) satisfies the condition PF + PG = 20.

Answer:

\(\frac{x^{2}}{100}\) + \(\frac{y^{2}}{64}\) = 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.

Answer:

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).

Answer:

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)

Answer:

\(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{64}\) = 1

b. An ellipse centered at the origin with x – intercepts ( – \(\sqrt{5}\), 0), (\(\sqrt{5}\), 0) and y – intercepts (0, 3), (0, – 3)

Answer:

\(\frac{x^{2}}{5}\) + \(\frac{y^{2}}{9}\) = 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.

Answer:

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?

Answer:

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

\(\frac{x^{2}}{4}\) + y^{2} = k.

a. k = 1

Answer:

b. k = \(\frac{1}{4}\)

Answer:

c. k = \(\frac{1}{9}\)

Answer:

d. k = \(\frac{1}{16}\)

Answer:

e. k = \(\frac{1}{25}\)

Answer:

f. k = \(\frac{1}{100}\)

Answer:

g. Make a conjecture: Which points in the plane satisfy the equation \(\frac{x^{2}}{4}\) + y^{2} = 0?

Answer:

As k is getting smaller, the ellipse is shrinking. It seems that the only point that lies on the curve given by \(\frac{x^{2}}{4}\) + y^{2} = 0 would be the single point (0, 0).

h. Explain why your conjecture in part (g) makes sense algebraically.

Answer:

Both \(\frac{x^{2}}{4}\) and y^{2} 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 y^{2} = 0, which means that (x, y) is (0, 0).

i. Which points in the plane satisfy the equation \(\frac{x^{2}}{4}\) + y^{2} = – 1?

Answer:

There are no points in the plane that satisfy the equation x^{2} + y^{2} = – 1 because \(\frac{x^{2}}{4}\) + y^{2} ≥ 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

\(\frac{x^{2}}{k}\) + y^{2} = 1.

a. k = 1

Answer:

b. k = 2

Answer:

c. k = 4

Answer:

d. k = 10

Answer:

e. k = 25

Answer:

f. Describe what happens to the graph of \(\frac{x^{2}}{k}\) + y^{2} = 1 as k→∞.

Answer:

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

x^{2} + \(\frac{y^{2}}{k}\) = 1.

a. k = 1

Answer:

b. k = 2

Answer:

c. k = 4

Answer:

d. k = 10

Answer:

e. k = 25

Answer:

f. Describe what happens to the graph of x^{2} + \(\frac{y^{2}}{k}\) = 1 as k → ∞.

Answer:

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.

Answer:

PF + PG = 4

\(\sqrt{(x – 1)^{2} + y^{2}}\) + \(\sqrt{(x + 1)^{2} + y^{2}}\) = 4

(x – 1)^{2} + y^{2} = 16 – 8\(\sqrt{(x + 1)^{2} + y^{2}}\) + (x + 1)^{2} + y^{2}

(x – 1)^{2} – (x + 1)^{2} – 16 = – 8\(\sqrt{(x + 1)^{2} + y^{2}}\)

x^{2} – 2x + 1 – (x^{2} + 2x + 1) – 16 = – 8\(\sqrt{(x + 1)^{2} + y^{2}}\)

– 4x – 16 = – 8\(\sqrt{(x + 1)^{2} + y^{2}}\)

x + 4 = 2\(\sqrt{(x + 1)^{2} + y^{2}}\)

x^{2} + 8x + 16 = 4(x^{2} + 2x + 1 + y^{2} )

x^{2} + 8x + 16 = 4x^{2} + 8x + 4 + 4y^{2}

– 3x^{2} – 4y^{2} = – 12

\(\frac{x^{2}}{4} + \frac{y^{2}}{3}\) = 1

b. Sketch the graph of the ellipse defined above.

Answer:

c. Verify that the x – intercepts of the graph satisfy the condition PF + PG = 4.

Answer:

For the x – intercepts (2, 0) and ( – 2, 0), we have PF + PG = 1 + 3 = 4 and PF + PG = 3 + 1 = 4.

d. Verify that the y – intercepts of the graph satisfy the condition PF + PG = 4.

Answer:

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.