Question

If $P$ and $Q$ are two points on the hyperbola $\frac{x^2}{5}-\frac{y^2}{8}=1$ whose centre is $C$ such that $CP$ is perpendicular to $CQ,$ then the value of $\frac{1}{(CP{)}^2}+\frac{1}{(CQ{)}^2}$ is equal to

• A. $\frac{3}{20}$
• B. $\frac{5}{40}$
• C. $\frac{13}{40}$
• D. $\frac{3}{40}$

Answer 1

The correct option is D $\frac{3}{40}$

Given hyperbola: $\frac{x^2}{5}-\frac{y^2}{8}=1\dots \left(1\right)$
Centre: $C(0,0)$
Let $P(x_1,y_1),Q(x_2,y_2)$ be two points on the hyperbola.
Let equation of $CP$ be $y=mx\dots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$ we get
$\frac{x^2}{5}-\frac{m^2{x}^2}{8}=1$
$\Rightarrow x^2=\frac{40}{8-5m^2}=x_1^2$
$\Rightarrow y^2=m^2{x}^2\Rightarrow \frac{40m^2}{8-5m^2}=y_1^2$

Now, $C{P}^2=x_1^2+y_1^2=\frac{40(1+m^2)}{8-5m^2}$
$\Rightarrow \frac{1}{C{P}^2}=\frac{8-5m^2}{40(1+m^2)}\dots \left(3\right)$
Similarly, $\frac{1}{C{Q}^2}=\frac{8m^2-5}{40(m^2+1)}\dots \left(4\right)$
$(\because CP\perp CQ\text{, replacing}m\text{with}\frac{-1}{m})$
$\therefore \frac{1}{C{P}^2}+\frac{1}{C{Q}^2}=\frac{3m^2+3}{40(m^2+1)}=\frac{3}{40}$