Question

$P$ is any point on the ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$ and $Q(0,2)$ and $R(0,-2)$ are two points. If $p_1$ and $p_2$ are the lengths of the perpendicular from $Q$ and $R$ on the tangent at $P$ then $\left(p_1^2+p_2^2\right)$ is equal to

• A. 32
• B. 16
• C. 4
• D. 8

Answer 1

Answer: (B)

Let the tangent be $m x-y+\sqrt{8 m^2+4}=0$
$p _1=\left|\frac{\sqrt{8 m ^2+4}-2}{\sqrt{1+ m ^2}}\right| ; p _2=\left|\frac{\sqrt{8 m ^2+4}+2}{\sqrt{1+ m ^2}}\right| $
$\Rightarrow\left( p _1^2+ p _2^2\right)=\frac{2\left(8 m ^2+4\right)+8}{\left(1+ m ^2\right)}=\frac{16\left(1+ m ^2\right)}{\left(1+ m ^2\right)}=16$