Q93 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
'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
- A6
- B7
- C8
- Correct answerD9
Solution
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 $=\left(\frac{0-4+C}{\sqrt{1+m^2}}\right)\left(\frac{0+4+C}{\sqrt{1+m^2}}\right)=\frac{C^2-16}{1+m^2}=\frac{9 m^2+25-16}{1+m^2}=9$
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 $=\left(\frac{0-4+C}{\sqrt{1+m^2}}\right)\left(\frac{0+4+C}{\sqrt{1+m^2}}\right)=\frac{C^2-16}{1+m^2}=\frac{9 m^2+25-16}{1+m^2}=9$
Hence, the correct answer is option: 'D'.