Conic Sections Question 369
Question: The equation of the normal to the hyperbola $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $ at the point $ (8,\ 3\sqrt{3}) $ is
[MP PET 1996]
Options:
A) $ \sqrt{3}x+2y=25 $
B) $ x+y=25 $
C) $ y+2x=25 $
D) $ 2x+\sqrt{3}y=25 $
Show Answer
Answer:
Correct Answer: D
Solution:
Applying the formula, the required normal is $ \frac{16x}{8}+\frac{9y}{3\sqrt{3}}=16+9i.e., $
$ 2x+\sqrt{3}y=25 $
Trick : This is the only equation among the given options at which the point $ (8,3\sqrt{3}) $ is located.
BETA
