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(D) We know that the product of the lengths of the perpendiculars drawn from the foci to any tangent of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}

Vedclass · Accepted Answer

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

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(D) We know that the product of the lengths of the perpendiculars drawn from the foci to any tangent of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is equal to $b^2$. Given that the product is $9$,we have $b^2 = 9$. The equation of the tangent is $y = \frac{-3}{4}x + 3\sqrt{2}$. Comparing this with the standard tangent equation $y = mx \pm \sqrt{a^2m^2 + b^2}$,where $m = \frac{-3}{4}$,we have $3\sqrt{2} = \sqrt{a^2(\frac{-3}{4})^2 + b^2}$. Squaring both sides,we get $18 = a^2(\frac{9}{16}) + 9$. Subtracting $9$ from both sides,$9 = \frac{9a^2}{16}$,which gives $a^2 = 16$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.

If the product of the lengths of the perpendiculars drawn from the foci to the tangent $y = \frac{-3}{4}x + 3\sqrt{2}$ of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $9$,then the eccentricity of that ellipse is

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