Conic Sections Question 122
The equation of the tangent to the hyperbola $ 4y^{2}=x^{2}-1 $ at the point (1, 0) is $ y = \frac{1}{2}x - \frac{1}{2} $
[Karnataka CET 1994]
Options:
A) $ x=1 $
B) $ y=1 $
C) $ y=4 $
D) $ x=4 $
Show Answer
Answer:
Correct Answer: A
Solution:
The equation of the tangent to $ 4y^{2}=x^{2}-1 $ at (1,0) is $ 4y= x $ or $ x-1=0 $ or $ x=1 $ .
BETA
