Let P be any point on the ellipse 7x^2 + 16y^ | vedclass

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(C) The given equation of the ellipse is $7x^2 + 16y^2 = 112$.Dividing by $112$,we get $\frac{x^2}{16} + \frac{y^2}{7} = 1$.Here,$a^2 = 16$ and $b^2 =

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$\frac{3}{4}$

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(C) The given equation of the ellipse is $7x^2 + 16y^2 = 112$.Dividing by $112$,we get $\frac{x^2}{16} + \frac{y^2}{7} = 1$.Here,$a^2 = 16$ and $b^2 = 7$.The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.$e = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.By the definition of an ellipse,the ratio of the distance of a point $P$ from the focus $S$ to its distance from the directrix $L$ is equal to the eccentricity $e$.Therefore,$\frac{SP}{PM} = e = \frac{3}{4}$.

Let $P$ be any point on the ellipse $7x^2 + 16y^2 = 112$,$S$ be a focus,$L$ be the corresponding directrix,and $PM$ be the perpendicular distance from $P$ to the directrix $L$. Then $\frac{SP}{PM} =$

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