The coordinates of the foci of the ellipse 16 | vedclass

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(B) Given the equation of the ellipse: $16x^{2} + 9y^{2} = 144$. Dividing both sides by $144$,we get: $\frac{16x^{2}}{144} + \frac{9y^{2}}{144} = 1$,w

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$(0, \pm \sqrt{7})$

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(B) Given the equation of the ellipse: $16x^{2} + 9y^{2} = 144$. Dividing both sides by $144$,we get: $\frac{16x^{2}}{144} + \frac{9y^{2}}{144} = 1$,which simplifies to $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$. Comparing this with the standard form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we have $a^{2} = 9$ and $b^{2} = 16$. Since $b^{2} > a^{2}$,the major axis is along the $y$-axis. Here,$a = 3$ and $b = 4$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$. The coordinates of the foci are $(0, \pm be) = (0, \pm 4 	imes \frac{\sqrt{7}}{4}) = (0, \pm \sqrt{7})$.

The coordinates of the foci of the ellipse $16x^{2} + 9y^{2} = 144$ are

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