Question

If $P$ and $Q$ are two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ whose center is $C$ such that $CP$ is perpendicular to $CQ,$ where $a<b$, then $\frac{1}{{CP}^2}+\frac{1}{{CQ}^2}$ is

• A. $-\frac{1}{a^2}+\frac{1}{b^2}$
• B. $\frac{1}{a^2}+\frac{1}{b^2}$
• C. $-\frac{1}{a^2}-\frac{1}{b^2}$
• D. $\frac{1}{a^2}-\frac{1}{b^2}$

Answer 1

The correct option is D $\frac{1}{a^2}-\frac{1}{b^2}$

Since the line $CP$ passes through the origin, i.e., centre, let its equation be $y=mx.$

The line $CP$ meets the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ in $P$ whose abscisse is given by

$\frac{x^2}{a^2}-\frac{m^2{x}^2}{b^2}=1x^2=\frac{a^2{b}^2}{b^2-a^2{m}^2}$

$\therefore y^2=m^2{x}^2=\frac{a^2{b}^2{m}^2}{b^2-a^2{m}^2}$

$\therefore {CP}^2=x^2+y^2=\frac{a^2{b}^2+a^2{b}^2{m}^2}{b^2-a^2{m}^2}$

Since $CQ\perp CP$ Replace $m$ by $-\frac{1}{m},$

we get ${CQ}^2=\frac{a^2{b}^2(1+\frac{1}{m^2})}{b^2-\frac{a^2}{m^2}}=\frac{a^2{b}^2(m^2+1)}{b^2{m}^2-a^2}$

$\therefore \frac{1}{{CP}^2}+\frac{1}{{CQ}^2}=\frac{b^2-a^2{m}^2+b^2{m}^2-a^2}{a^2{b}^2(1+m^2)}=\frac{b^2-a^2}{a^2{b}^2}=\frac{1}{a^2}-\frac{1}{b^2}.$