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: 33

Answer

$ \quad \displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

Work Step by Step

The center is at the midpoint of the vertices, $(0,0)$ The foci, center and vertices lie on the same line, the main axis. Main axis here: $y=0$ (the $x$-axis). Table 3: Major axis horizontal (parallel to the x-axis)$\begin{array}{lll} \text{ Foci}&\text{ Vertices}&\text{ Equation}\\ {(h+c, k)}&{(h+a,k)}&{\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1}\\ {(h-c,k)}&{(h-a,k)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\\ \end{array}$ --- Vertices at $(\pm 5,0) \quad \Rightarrow \quad a=5$ Focus at $(-4,0) \quad \Rightarrow \quad c=4$ Find b: $b^{2}=a^{2}-c^{2}=25-16=9$ $b=3$ The equation is $ \quad \displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
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