Conic Sections Question 36
Question: The equation of the tangent at the point $ (a\sec \theta ,\ b\tan \theta ) $ of the conic $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ , is
Options:
A) $ x{{\sec }^{2}}\theta -y{{\tan }^{2}}\theta =1 $
B) $ \frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1 $
C) $ \frac{x+a\sec \theta }{a^{2}}-\frac{y+b\tan \theta }{b^{2}}=1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{x(a\sec \theta )}{a^{2}}-\frac{y(b\tan \theta )}{b^{2}}=1 $
therefore $ \frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1 $ .
BETA
