Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - 10.3 The Ellipse - 10.3 Assess Your Understanding - Page 657: 24


Vertices: $(0,-3),(0,3)$ Foci: $(0,-\sqrt{5}),(0,\sqrt{5})$ See graph

Work Step by Step

We are given the ellipse: $4y^2+9x^2=36$ Put the equation in standard form: $\dfrac{4y^2}{36}+\dfrac{9x^2}{36}=1$ $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$ The equation is in the form: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k,a,b,c$: $h=0$ $k=0$ $a^2=9\Rightarrow a=\sqrt{9}=3$ $b^2=4\Rightarrow b=\sqrt 4=2$ $c^2=a^2-b^2=9-4=5$ $c=\sqrt{5}$ The center is: $(h,k)=(0,0)$ Find the vertices: $(h,k-a)=(0,0-3)=(0,-3)$ $(h,k+a)=(0,0+3)=(0,3)$ Find the foci: $(h,k-c)=(0,0-\sqrt{5})=(0,-\sqrt{5})$ $(h,k+c)=(0,0+\sqrt{5})=(0,\sqrt{5})$ Use the value of $b=2$ to find the two points to the left and right of the center: $(h-b,k)=(0-2,0)=(-2,0)$ $(h+b,k)=(0+2,0)=(2,0)$ Plot the center, the vertices, the two points to the left and the right of the center, and the foci. Then, graph the ellipse: Vertices: $(0,-3),(0,3)$ Foci: $(0,-\sqrt{5}),(0,\sqrt{5})$ See graph
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