Question

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

• 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 abscissa is given by

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

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

$\therefore {CP}^2=x^2+y^2$

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

Since $CQ\perp CP$

Replace $m$ by $-\frac{1}{m}$, we get

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

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

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

Hence, option D.