'The product of 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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Aicharya · Accepted Answer

Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{25}=1$, Here $b \gt a$ $\begin{aligned} & \therefore c^2=b^2-a^2=25-9=16 \Rightarrow c= \pm 4 \\ & \text { foci }=(0, \pm 4) \end{aligned}$
Let the tangent be $y=m x+C$ where $C^2=a^2 m^2+b^2 \Rightarrow C^2=9 m^2+25$...(i)
Product of distance from focii to tangent $=\left(\frac{0-4+C}{\sqrt{1+m^2}}\right)\left(\frac{0+4+C}{\sqrt{1+m^2}}\right)=\frac{C^2-16}{1+m^2}=\frac{9 m^2+25-16}{1+m^2}=9$

'The product of perpendiculars from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ on the…

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