Conic Sections Question 216
Question: Let $ P(asec\theta ,btan\theta ) $ and Q $ Q(asec\phi ,btan\phi ), $ where $ \theta +\phi =\pi /2, $ be two points on the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1. $ If $ (h,k) $ is the point of intersection of the normal at P and Q, then kz is equal to
Options:
A) $ \frac{a^{2}+b^{2}}{a} $
B) $ -( \frac{a^{2}+b^{2}}{a} ) $
C) $ \frac{a^{2}+b^{2}}{b} $
D) $ -( \frac{a^{2}+b^{2}}{b} ) $
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Answer:
Correct Answer: D
Solution:
[d] Equation of the normal to the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ at the point $ (aseca,btana) $ is given by $ ax\cos \alpha +by\cot \alpha =a^{2}+b^{2} $ Normal at $ \theta ,\phi $ are $ \begin{matrix} ax\cos \theta +by\cot \theta =a^{2}+b^{2} \\ ax\cos \phi +by\cot \phi =a^{2}+b^{2} \\ \end{matrix} . $ where $ \phi =\frac{\pi }{2}-\theta $ and these passes through $ (h,k) $
$ \therefore ah\cos \theta +bk\cot \theta =a^{2}+b^{2} $
$ \Rightarrow ah\sin \theta +bk\tan \theta =a^{2}+b^{2} $ Eliminating $ h,bk(cot\theta sin\theta -tan\theta cos\theta ) $
$ =(a^{2}+b^{2})(sin\theta -cos\theta ) $ or $ k=-(a^{2}+b^{2})/b $
BETA
