Conic Sections Question 183
Question: If the straight line $ x\cos \alpha +y\sin \alpha =p $ be a tangent to the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ , then
[Karnataka CET 1999]
Options:
A) $ a^{2}{{\cos }^{2}}\alpha +b^{2}{{\sin }^{2}}\alpha =p^{2} $
B) $ a^{2}{{\cos }^{2}}\alpha -b^{2}{{\sin }^{2}}\alpha =p^{2} $
C) $ a^{2}{{\sin }^{2}}\alpha +b^{2}{{\cos }^{2}}\alpha =p^{2} $
D) $ a^{2}{{\sin }^{2}}\alpha -b^{2}{{\cos }^{2}}\alpha =p^{2} $
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Answer:
Correct Answer: B
Solution:
$ x\cos \alpha +y\sin \alpha =p\Rightarrow y=-\cot \alpha .x+p\text{cosec }\alpha $
It is tangent to the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $
Therefore, $ p^{2}cose{c^{2}}\alpha =a^{2}{{\cot }^{2}}\alpha -b^{2} $
$ \Rightarrow a^{2}{{\cos }^{2}}\alpha -b^{2}{{\sin }^{2}}\alpha =p^{2} $ .
BETA
