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$C$ the centre of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$. The tangents at any point $P$ on this hyperbola meets the straight lines $bx - ay = 0$ and $bx + ay = 0$ in the points $Q$ and $R$ respectively. Then $CQ\;.\;CR = $
${a^2} + {b^2}$
${a^2} - {b^2}$
$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$
$\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}}$
Solution
(a) $P$ is $(a\sec \theta ,b\tan \theta )$
Tangen $t$ at $P$ is $\frac{{x\sec \theta }}{a} – \frac{{y\tan \theta }}{b} = 1$
It meets $bx – ay = 0$
$i.e.$, $\frac{x}{a} = \frac{y}{b}$ in $Q$
$ Q$ is $\left( {\frac{a}{{\sec \theta – \tan \theta }},\frac{{ – b}}{{\sec \theta – \tan \theta }}} \right)$
It meets $bx + ay = 0$
$i.e.$, $\frac{x}{a} = – \frac{y}{b}$ in $R.$
$ R$ is $\left( {\frac{a}{{\sec \theta + \tan \theta }},\frac{{ – b}}{{\sec \theta + \tan \theta }}} \right)$
$CQ.CR = \frac{{\sqrt {{a^2} + {b^2}} }}{{(\sec \theta – \tan \theta )}}.\frac{{\sqrt {{a^2} + {b^2}} }}{{(\sec \theta + \tan \theta )}}$
$ = {a^2} + {b^2}$, $\{ \because {\sec ^2}\theta – {\tan ^2}\theta = 1\} $ .
