If a focal chord of the ellipse frac{x^2}{25} | vedclass

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(D) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.Here,$a^2=25$ and $b^2=16$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{

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$\frac{3}{\sqrt{2}}$

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(D) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.Here,$a^2=25$ and $b^2=16$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The foci are $(\pm ae, 0) = (\pm 5 	imes \frac{3}{5}, 0) = (\pm 3, 0)$.Let the focal chord pass through the focus $F(3,0)$ and the point $A(0,3)$ on the minor axis.The slope of the chord $AF$ is $m = \frac{3-0}{0-3} = -1$.The equation of the chord is $y - 0 = -1(x - 3)$,which simplifies to $x + y - 3 = 0$.The perpendicular distance $d$ from the centre $(0,0)$ to the line $x + y - 3 = 0$ is given by $d = \frac{|0 + 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$.

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,then the perpendicular distance from the centre of the ellipse to this focal chord is

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