Find the equation of the ellipse whose vertic | vedclass

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(A) Since the vertices are on the $x$-axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a$ is the

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$\frac{x^{2}}{169} + \frac{y^{2}}{144} = 1$

Vedclass · Answer

(A) Since the vertices are on the $x$-axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a$ is the semi-major axis.Given that the vertices are $(\pm 13, 0)$,we have $a = 13$.The foci are $(\pm 5, 0)$,so $c = 5$.Using the relation $c^{2} = a^{2} - b^{2}$,we get:$5^{2} = 13^{2} - b^{2}$$25 = 169 - b^{2}$$b^{2} = 169 - 25 = 144$,which implies $b = 12$.Substituting the values of $a^{2}$ and $b^{2}$ into the standard equation,we get $\frac{x^{2}}{169} + \frac{y^{2}}{144} = 1$.

Find the equation of the ellipse whose vertices are $(\pm 13, 0)$ and foci are $(\pm 5, 0)$.

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