Conic Sections Question 429
Question: The equation of the normal to the hyperbola $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $ at $ (-4,\ 0) $ is
[UPSEAT 2002]
Options:
A) $ y=0 $
B) $ y=x $
C) $ x=0 $
D) $ x=-y $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $
$ \Rightarrow $ $ \frac{2x}{16}-\frac{2y}{9}\frac{dy}{dx}=0 $
therefore $ \frac{dy}{dx}=\frac{2x\times 9}{16\times 2y} $
$ =\frac{9}{16}\frac{x}{y} $
therefore $ {{( \frac{-dx}{dy} )} _{(-4,0)}}=\frac{-16}{9}\frac{y}{x}=0 $
Hence, equation of normal
therefore $ (y-0)=0(x+4) $
therefore $ y=0. $
BETA
