College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 7 - Section 7.3 - The Ellipse - 7.3 Assess Your Understanding - Page 525: 24

Answer

Center at $\quad(0,0).$ Vertices at $\quad(0,\pm 3).$ Foci at $\quad(0,\pm\sqrt{5})$

Work Step by Step

Divide the equation with $36$ ... (we want the standard form) $\displaystyle \frac{4y^{2}}{36}+\frac{9x^{2}}{36}=1$ $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ $\displaystyle \frac{x^{2}}{2^{2}}+\frac{y^{2}}{3^{2}}=1,\quad$ main axis: $y$-axis$, a=3,\ b=2.$ Table 3: Major axis vertical (parallel to the y-axis) $\begin{array}{lll} \text{ Foci}&\text{ Vertices}&\text{ Equation}\\ {(h,k+c)}&{(h,k+a)}&{\displaystyle \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\\ {(h,k-c)}&{(h,k-a)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\end{array}$ The center of the ellipse is at the origin. The vertices are $(0,3)$ and $(0,-3).$ Find $c:$ $c^{2}=a^{2}-b^{2}$ $c^{2}=9-4$ $c^{2}=5\quad $ $c=\sqrt{5}\approx 2.236$ The foci are $(0,\sqrt{5})$ and $(0,-\sqrt{5})$
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