If two points P and Q on the hyperbola frac{x | vedclass

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(B) Let the equation of line $CP$ be $y = mx$. Substituting this into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $\frac{x^2

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

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(B) Let the equation of line $CP$ be $y = mx$. Substituting this into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $\frac{x^2}{a^2} - \frac{m^2 x^2}{b^2} = 1$.Solving for $x^2$,we have $x^2 = \frac{a^2 b^2}{b^2 - a^2 m^2}$.Since $y^2 = m^2 x^2$,we have $y^2 = \frac{a^2 b^2 m^2}{b^2 - a^2 m^2}$.Then $(CP)^2 = x^2 + y^2 = x^2(1 + m^2) = \frac{a^2 b^2(1 + m^2)}{b^2 - a^2 m^2}$.Since $CP \perp CQ$,the slope of $CQ$ is $-\frac{1}{m}$. Replacing $m$ with $-\frac{1}{m}$ in the expression for $(CP)^2$,we get $(CQ)^2 = \frac{a^2 b^2(1 + (-1/m)^2)}{b^2 - a^2(-1/m)^2} = \frac{a^2 b^2(1 + m^2)/m^2}{(b^2 m^2 - a^2)/m^2} = \frac{a^2 b^2(1 + m^2)}{b^2 m^2 - a^2}$.Now,$\frac{1}{(CP)^2} + \frac{1}{(CQ)^2} = \frac{b^2 - a^2 m^2}{a^2 b^2(1 + m^2)} + \frac{b^2 m^2 - a^2}{a^2 b^2(1 + m^2)} = \frac{b^2 - a^2 m^2 + b^2 m^2 - a^2}{a^2 b^2(1 + m^2)} = \frac{b^2(1 + m^2) - a^2(1 + m^2)}{a^2 b^2(1 + m^2)} = \frac{b^2 - a^2}{a^2 b^2} = \frac{1}{a^2} - \frac{1}{b^2}$.

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

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