Conic Sections Question 331
Question: The equation of the normal at the point $ (a\sec \theta ,\ b\tan \theta ) $ of the curve $ b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2} $ is
[Karnataka CET 1999]
Options:
A) $ \frac{ax}{\cos \theta }+\frac{by}{\sin \theta }=a^{2}+b^{2} $
B) $ \frac{ax}{\tan \theta }+\frac{by}{\sec \theta }=a^{2}+b^{2} $
C) $ \frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^{2}+b^{2} $
D) $ \frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^{2}-b^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of normal to hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ at $ (a\sec \theta ,b\tan \theta ) $ is $ \frac{a^{2}x}{a\sec \theta }+\frac{b^{2}y}{b\tan \theta }=a^{2}+b^{2} $ .
BETA
