Applications Of Derivatives Question 137
Question: The equation of tangent at $ (-4,,-4) $ on the curve $ x^{2}=-4y $ is
[Karnataka CET 2001: Pb. CET 2000]
Options:
A) $ 2x+y+4=0 $
B) $ 2x-y-12=0 $
C) $ 2x+y-4=0 $
D) $ 2x-y+4=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ x^{2}=-4y $
therefore $ 2x=-4\frac{dy}{dx} $
therefore $ \frac{dy}{dx}=\frac{-x}{2} $
therefore $ {{( \frac{dy}{dx} )} _{(-4,,-4)}}=2 $ .
We know that equation of tangent is, $ (y-y_1),=,{{( \frac{dy}{dx} )} _{(x_1,,y_1)}}(x-x_1) $
therefore $ y+4=2(x+4) $
therefore $ 2x-y+4=0 $ .
BETA
