{"id":101227,"date":"2022-07-12T11:32:46","date_gmt":"2022-07-12T06:02:46","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=101227"},"modified":"2022-05-10T14:58:03","modified_gmt":"2022-05-10T09:28:03","slug":"eureka-math-precalculus-module-3-lesson-7","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-precalculus-module-3-lesson-7\/","title":{"rendered":"Eureka Math Precalculus Module 3 Lesson 7 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Precalculus Module 3 Lesson 7 Answer Key<\/h2>\n

Eureka Math Precalculus Module 3 Lesson 7 Exercise Answer Key<\/h3>\n

Exercise
\nPoints 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.
\n\"Engage
\nAnswer:
\nThe 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.
\nPF + PG = \\(\\sqrt{x^{2} + (3 – y)^{2}}\\) + \\(\\sqrt{x^{2} + (y + 3)^{2}}\\) = 8
\n\\(\\sqrt{x^{2} + (3 – y)^{2}}\\) = 8 – \\(\\sqrt{x^{2} + (y + 3)^{2}}\\)
\nx2<\/sup> + (3 – y)2<\/sup> = 64 – 16\\(\\sqrt{x^{2} + (y + 3)^{2}}\\) + x2<\/sup> + (y + 3)2<\/sup>
\ny2<\/sup> – 6y + 9 = 64 – 16\\(\\sqrt{x^{2} + (y + 3)^{2}}\\) + y2<\/sup> + 6y + 9
\n16\\(\\sqrt{x^{2} + (y + 3)^{2}}\\) = 64 + 12y
\n256[x2<\/sup> + (y + 3)2<\/sup> ] = 4096 + 1536y + 144y2<\/sup>
\n256[x2<\/sup> + y2<\/sup> + 6y + 9] = 4096 + 1536y + 144y2<\/sup>
\n256x2<\/sup> + 256y2<\/sup> + 1536y + 2304 = 4096 + 1536y + 144y2<\/sup>
\n256x2<\/sup> + 112y2<\/sup> = 1792
\n\\(\\frac{x^{2}}{7} + \\frac{y^{2}}{16}\\) = 1<\/p>\n

Eureka Math Precalculus Module 3 Lesson 7 Problem Set Answer Key<\/h3>\n

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

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

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

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.
\nAnswer:
\n\\(\\frac{x^{2}}{\\frac{9}{4}}\\) + \\(\\frac{y^{2}}{\\frac{65}{36}}\\) = 1
\n\\(\\frac{4x^{2}}{9}\\) + \\(\\frac{36y^{2}}{65}\\) = 1<\/p>\n

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

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

Question 2.
\nRecall 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.
\nAnswer:
\na. a = \\(\\frac{5}{2}\\), b = \\(\\frac{3}{2}\\)
\nb. a = \\(\\frac{5}{2}\\), b = \\(\\frac{\\sqrt{21}}{2}\\)
\nc. a = \\(\\sqrt{3}\\), b = 2
\nd. a = \\(\\frac{3}{2}\\), b = \\(\\frac{\\sqrt{65}}{6}\\)
\ne. a = \\(\\sqrt{11}\\), b = 6
\nf. a = 10, b = 8<\/p>\n

Question 3.
\nSummarize what you know about equations of ellipses centered at the origin with vertices (a, 0), ( – a, 0), (0, b), and (0, – b).
\nAnswer:
\nFor 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
\nx – 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.<\/p>\n

Question 4.
\nUse your answer to Problem 3 to find the equation of the ellipse for each of the situations below.
\na. An ellipse centered at the origin with x – intercepts ( – 2, 0), (2, 0) and y – intercepts (0, 8), (0, – 8)
\nAnswer:
\n\\(\\frac{x^{2}}{4}\\) + \\(\\frac{y^{2}}{64}\\) = 1<\/p>\n

b. An ellipse centered at the origin with x – intercepts ( – \\(\\sqrt{5}\\), 0), (\\(\\sqrt{5}\\), 0) and y – intercepts (0, 3), (0, – 3)
\nAnswer:
\n\\(\\frac{x^{2}}{5}\\) + \\(\\frac{y^{2}}{9}\\) = 1<\/p>\n

Question 5.
\nExamine 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.
\nAnswer:
\nIf 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.<\/p>\n

Question 6.
\nIs 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?
\nAnswer:
\nNo. Since 9 > 4, the foci must be along the y – axis.<\/p>\n

Question 7.
\nFor each value of k specified in parts (a)\u2013(e), plot the set of points in the plane that satisfy the equation
\n\\(\\frac{x^{2}}{4}\\) + y2<\/sup> = k.
\na. k = 1
\nAnswer:
\n\"Eureka<\/p>\n

b. k = \\(\\frac{1}{4}\\)
\nAnswer:
\n\"Eureka<\/p>\n

c. k = \\(\\frac{1}{9}\\)
\nAnswer:
\n\"Eureka<\/p>\n

d. k = \\(\\frac{1}{16}\\)
\nAnswer:
\n\"Eureka<\/p>\n

e. k = \\(\\frac{1}{25}\\)
\nAnswer:
\n\"Eureka<\/p>\n

f. k = \\(\\frac{1}{100}\\)
\nAnswer:
\n\"Eureka<\/p>\n

g. Make a conjecture: Which points in the plane satisfy the equation \\(\\frac{x^{2}}{4}\\) + y2<\/sup> = 0?
\nAnswer:
\nAs 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<\/sup> = 0 would be the single point (0, 0).<\/p>\n

h. Explain why your conjecture in part (g) makes sense algebraically.
\nAnswer:
\nBoth \\(\\frac{x^{2}}{4}\\) and y2<\/sup> 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<\/sup> = 0, which means that (x, y) is (0, 0).<\/p>\n

i. Which points in the plane satisfy the equation \\(\\frac{x^{2}}{4}\\) + y2<\/sup> = – 1?
\nAnswer:
\nThere are no points in the plane that satisfy the equation x2<\/sup> + y2<\/sup> = – 1 because \\(\\frac{x^{2}}{4}\\) + y2<\/sup> \u2265 0 for all real numbers x and y.<\/p>\n

Question 8.
\nFor each value of k specified in parts (a)\u2013(e), plot the set of points in the plane that satisfy the equation
\n\\(\\frac{x^{2}}{k}\\) + y2<\/sup> = 1.
\na. k = 1
\nAnswer:
\n\"Eureka<\/p>\n

b. k = 2
\nAnswer:
\n\"Eureka<\/p>\n

c. k = 4
\nAnswer:
\n\"Eureka<\/p>\n

d. k = 10
\nAnswer:
\n\"Eureka<\/p>\n

e. k = 25
\nAnswer:
\n\"Eureka<\/p>\n

f. Describe what happens to the graph of \\(\\frac{x^{2}}{k}\\) + y2<\/sup> = 1 as k\u2192\u221e.
\nAnswer:
\nAs 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).<\/p>\n

Question 9.
\nFor each value of k specified in parts (a)\u2013(e), plot the set of points in the plane that satisfy the equation
\nx2<\/sup> + \\(\\frac{y^{2}}{k}\\) = 1.
\na. k = 1
\nAnswer:
\n\"Eureka<\/p>\n

b. k = 2
\nAnswer:
\n\"Eureka<\/p>\n

c. k = 4
\nAnswer:
\n\"Eureka<\/p>\n

d. k = 10
\nAnswer:
\n\"Eureka<\/p>\n

e. k = 25
\nAnswer:
\n\"Eureka<\/p>\n

f. Describe what happens to the graph of x2<\/sup> + \\(\\frac{y^{2}}{k}\\) = 1 as k \u2192 \u221e.
\nAnswer:
\nAs 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}\\)).<\/p>\n

Eureka Math Precalculus Module 3 Lesson 7 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nSuppose that the foci of an ellipse are F( – 1, 0) and G(1, 0) and that the point P(x, y) satisfies the condition
\nPF + PG = 4.
\na. 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.
\nAnswer:
\nPF + PG = 4
\n\\(\\sqrt{(x – 1)^{2} + y^{2}}\\) + \\(\\sqrt{(x + 1)^{2} + y^{2}}\\) = 4
\n(x – 1)2<\/sup> + y2<\/sup> = 16 – 8\\(\\sqrt{(x + 1)^{2} + y^{2}}\\) + (x + 1)2<\/sup> + y2<\/sup>
\n(x – 1)2<\/sup> – (x + 1)2<\/sup> – 16 = – 8\\(\\sqrt{(x + 1)^{2} + y^{2}}\\)
\nx2<\/sup> – 2x + 1 – (x2<\/sup> + 2x + 1) – 16 = – 8\\(\\sqrt{(x + 1)^{2} + y^{2}}\\)
\n– 4x – 16 = – 8\\(\\sqrt{(x + 1)^{2} + y^{2}}\\)
\nx + 4 = 2\\(\\sqrt{(x + 1)^{2} + y^{2}}\\)
\nx2<\/sup> + 8x + 16 = 4(x2<\/sup> + 2x + 1 + y2<\/sup> )
\nx2<\/sup> + 8x + 16 = 4x2<\/sup> + 8x + 4 + 4y2<\/sup>
\n– 3x2<\/sup> – 4y2<\/sup> = – 12
\n\\(\\frac{x^{2}}{4} + \\frac{y^{2}}{3}\\) = 1<\/p>\n

b. Sketch the graph of the ellipse defined above.
\n\"Eureka
\nAnswer:
\n\"Eureka<\/p>\n

c. Verify that the x – intercepts of the graph satisfy the condition PF + PG = 4.
\nAnswer:
\nFor the x – intercepts (2, 0) and ( – 2, 0), we have PF + PG = 1 + 3 = 4 and PF + PG = 3 + 1 = 4.<\/p>\n

d. Verify that the y – intercepts of the graph satisfy the condition PF + PG = 4.
\nAnswer:
\nFor 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.<\/p>\n","protected":false},"excerpt":{"rendered":"

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 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/101227"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=101227"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/101227\/revisions"}],"predecessor-version":[{"id":136188,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/101227\/revisions\/136188"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=101227"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=101227"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=101227"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}