• # question_answer 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)            ${{a}^{2}}+{{b}^{2}}$               B)            ${{a}^{2}}-{{b}^{2}}$ C)            $\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}$                                    D)            $\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}$

Correct Answer: A

Solution :

$P$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 $\left( \frac{a}{\sec \theta +\tan \theta },\frac{-b}{\sec \theta +\tan \theta } \right)$            \ $CQ.CR=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{(\sec \theta -\tan \theta )}.\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{(\sec \theta +\tan \theta )}$                                  $={{a}^{2}}+{{b}^{2}}$,  $\{\because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\}$.

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