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 $\frac{x^{2}}{36}+\frac{y^{2}}{20}=1$.

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