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If two points $P$ and $Q$ on the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ whose centre is $C$, are such that $CP$ is perpendicular to $CQ, ( a < b )$ , then value of, $\frac{1}{{{{(CP)}^2}}} + \frac{1}{{{{(CQ)}^2}}} = $
$\frac {1}{ab}$
$\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}}$
$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$
$\frac{1}{{{a^2} + {b^2}}}$
Solution
Let equation of $\mathrm{CP}$ is $\mathrm{y}=\mathrm{mx}$ and
$\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1$
$\frac{x^{2}}{a^{2}}-\frac{m^{2} x^{2}}{b^{2}}=1$
After solving, $x^{2}=\frac{a^{2} b^{2}}{b^{2}-a^{2} m^{2}}$ and
$y^{2}=\frac{a^{2} m^{2} b^{2}}{b^{2}-a^{2} m^{2}}$
For $\mathrm{CQ},$ equation of line is $\mathrm{y}=\frac{-1}{\mathrm{m}} \mathrm{x}$
Again by solving $\mathrm{CQ}^{2}=\frac{\mathrm{a}^{2} \mathrm{b}^{2}\left(1+\mathrm{m}^{2}\right)}{\mathrm{b}^{2} \mathrm{m}^{2}-\mathrm{a}^{2}}$
So, $\frac{1}{\mathrm{CP}^{2}}+\frac{1}{\mathrm{CQ}^{2}}=\frac{\mathrm{b}^{2}\left(1+\mathrm{m}^{2}\right)-\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}{\mathrm{a}^{2} \mathrm{b}^{2}\left(1+\mathrm{m}^{2}\right)}$
$=\frac{1}{\mathrm{a}^{2}}-\frac{1}{\mathrm{b}^{2}}$
