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

Question

Aicharya · Accepted Answer

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 $=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\sqrt{\frac{18}{4}}$ $=\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}$

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,…

Solution