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: 35

Answer

$\quad \displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{13}=1$
1563832151

Work Step by Step

The center is at the midpoint of the foci, $(0,0)$ The foci, center and vertices lie on the same line, the main axis. Main axis here: $x=0$ (the $y$-axis). Table 3: Major axis vertical (parallel to the y-axis) $\begin{array}{lll} \text{ Foci}&\text{ Vertices}&\text{ Equation}\\ {(h,k+c)}&{(h,k+a)}&{\displaystyle \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\\ {(h,k-c)}&{(h,k-a)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\end{array}$ --- $x$ -intercepts are $\pm 2 \quad \Rightarrow \quad b=2$ Foci at $(0,\pm 3) \quad \Rightarrow \quad c=3$ Find $a:$ $a^{2}=b^{2}+c^{2}=13$ $a=\sqrt{13}\approx 3.606$ The equation is $\quad \displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{13}=1$
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