Conic Sections Question 196
Question: If the tangent on the point $ (2\sec \varphi ,\ 3\tan \varphi ) $ of the hyperbola $ \frac{x^{2}}{4}-\frac{y^{2}}{9}=1 $ is parallel to $ 3x-y+4=0 $ , then the value of f is
[MP PET 1998]
Options:
A) $ 45^{o} $
B) $ 60^{o} $
C) $ 30^{o} $
D) $ 75^{o} $
Show Answer
Answer:
Correct Answer: C
Solution:
Differentiation of $ x=2\sec \varphi $
therefore $ \frac{dx}{d\varphi }=2\sec \varphi \tan \varphi $ Differentiate, $ y=3\tan \varphi $ w.r.t. f, we get $ \frac{dy}{d\varphi }=3{{\sec }^{2}}\varphi $
$ \therefore $ Gradient of tangent $ \frac{dy}{dx}=\frac{dy/d\varphi }{dx/d\varphi }=\frac{3{{\sec }^{2}}\varphi }{2\sec \varphi \tan \varphi } $
$ \frac{dy}{dx}=\frac{3}{2}cosec\varphi $ ……(i) But, tangent is parallel to $ 3x-y+4=0 $
$ \therefore $ Gradient $ m=3 $ ……(ii) By (i) and (ii), $ \frac{3}{2}cosec\varphi =3 $
therefore $ cosec\varphi =2 $ ,
$ \therefore \varphi =30{}^\circ $ .
BETA
