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 $mx-y+\sqrt{8m^2+4}=0$
$p_1=\left|\frac{\sqrt{8m^2+4}-2}{\sqrt{1+m^2}}\right|;p_2=\left|\frac{\sqrt{8m^2+4}+2}{\sqrt{1+m^2}}\right|$
$\Rightarrow \left(p_1^2+p_2^2\right)=\frac{2\left(8m^2+4\right)+8}{\left(1+m^2\right)}=\frac{16\left(1+m^2\right)}{\left(1+m^2\right)}=16$