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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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The correct answer is:
$\frac{3}{\sqrt{2}}$
Let $A=(0,3)$
Given the ellipse, $\frac{x^2}{25}+\frac{y^2}{16}=1$
So focus of the ellipse $F=\left(\sqrt{5^2-4^2}, 0\right)$
$\Rightarrow \mathrm{F}=(3,0), \mathrm{P}\left(\frac{0+3}{2}, \frac{3+0}{2}\right)=\left(\frac{3}{2}, \frac{3}{2}\right)$
So required distance $=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\sqrt{\frac{18}{4}}$
$=\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}$
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