If C is the centre of the ellipse 9x^2 + 16y^ | vedclass

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(D) The given equation of the ellipse is $9x^2 + 16y^2 = 144$.Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Here,$a^2 = 16$ and $b^2 =

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$\sqrt{7} : 8$

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(D) The given equation of the ellipse is $9x^2 + 16y^2 = 144$.Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Here,$a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.The centre $C$ is $(0, 0)$ and the focus $S$ is $(ae, 0) = (4 	imes \frac{\sqrt{7}}{4}, 0) = (\sqrt{7}, 0)$.The distance $CS = ae = \sqrt{7}$.The length of the major axis is $2a = 2 	imes 4 = 8$.Therefore,the ratio $CS : 2a = \sqrt{7} : 8$.

If $C$ is the centre of the ellipse $9x^2 + 16y^2 = 144$ and $S$ is one focus,then the ratio of $CS$ to the major axis is:

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