Find the equation for the ellipse that satisf | vedclass

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(A) The vertices are given as $(0, \pm 13)$,which lie on the $y$-axis. Thus,the major axis is along the $y$-axis,and the equation of the ellipse is of

Vedclass · Accepted Answer

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

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(A) The vertices are given as $(0, \pm 13)$,which lie on the $y$-axis. Thus,the major axis is along the $y$-axis,and the equation of the ellipse is of the form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$. Here,$a = 13$ and the foci are $(0, \pm c)$ where $c = 5$. Using the relation $a^{2} = b^{2} + c^{2}$,we have $13^{2} = b^{2} + 5^{2}$. $169 = b^{2} + 25$. $b^{2} = 169 - 25 = 144$. Substituting the values of $a^{2}$ and $b^{2}$ into the standard equation,we get $\frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Vertices $(0, \pm 13)$,foci $(0, \pm 5)$.

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