The product of the perpendiculars from the tw | vedclass

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(D) The given equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$. Here,$b^2 = 25$ and $a^2 = 9$,so $b > a$.The foci are located at $(0, \

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$9$

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(D) The given equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$. Here,$b^2 = 25$ and $a^2 = 9$,so $b > a$.The foci are located at $(0, \pm c)$,where $c^2 = b^2 - a^2 = 25 - 9 = 16$,so $c = 4$. Thus,the foci are $S_1(0, 4)$ and $S_2(0, -4)$.Let the equation of the tangent to the ellipse be $y = mx + C$,where $C^2 = a^2m^2 + b^2 = 9m^2 + 25$.The perpendicular distance $p_1$ from $(0, 4)$ to $mx - y + C = 0$ is $p_1 = \frac{|-4 + C|}{\sqrt{m^2 + 1}}$.The perpendicular distance $p_2$ from $(0, -4)$ to $mx - y + C = 0$ is $p_2 = \frac{|4 + C|}{\sqrt{m^2 + 1}}$.The product of the perpendiculars is $p_1 p_2 = \frac{|C^2 - 16|}{m^2 + 1} = \frac{9m^2 + 25 - 16}{m^2 + 1} = \frac{9m^2 + 9}{m^2 + 1} = \frac{9(m^2 + 1)}{m^2 + 1} = 9$.

The product of the perpendiculars from the two foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ on the tangent at any point on the ellipse is:

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