The eccentricity of the ellipse given by the | vedclass

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

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$\frac{\sqrt{7}}{4}$

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(A) The given equation of the ellipse is $9x^{2} + 16y^{2} = 144$. Dividing both sides by $144$,we get: $\frac{9x^{2}}{144} + \frac{16y^{2}}{144} = 1$ $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$. Comparing this with the standard equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we have $a^{2} = 16$ and $b^{2} = 9$. Since $a^{2} > b^{2}$,the eccentricity $e$ is given by the formula: $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{16 - 9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.

The eccentricity of the ellipse given by the equation $9x^{2} + 16y^{2} = 144$ is

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