Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0)
Question:
Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6,0)$, foci $(\pm 4,0)$
Solution:
Vertices $(\pm 6,0)$, foci $(\pm 4,0)$
Here, the vertices are on the $x$-axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where a is the semi-major axis.
Accordingly, a = 6, c = 4.
It is known that $a^{2}=b^{2}+c^{2}$.
$\therefore 6^{2}=b^{2}+4^{2}$
$\Rightarrow 36=b^{2}+16$
$\Rightarrow b^{2}=36-16$
$\Rightarrow b=\sqrt{20}$
Thus, the equation of the ellipse is $\frac{x^{2}}{6^{2}}+\frac{y^{2}}{(\sqrt{20})^{2}}=1$ or $\frac{x^{2}}{36}+\frac{y^{2}}{20}=1$.