Applications Of Derivatives Question 29
Question: The equation of the normal to the curve $ y=| x^{2}-| x | | $ at $ x=-2 $ .
Options:
A) $ 3y=x+8 $
B) $ x=3y+4 $
C) $ y=2x+8 $
D) $ y=3x $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] In the neighborhood of $ x=-2,y=x^{2}+x $ .
Hence, the point on curve is $ (-2,2) $ . $ \frac{dy}{dx}=2x+1 $ or $ {{. \frac{dy}{dx} |}_{x=-2}}=-3 $ So, the slope of the normal at $ (-2,2) $ is $ \frac{1}{3} $ .
Hence, the equation of the normal is $ \frac{1}{3}(x+2)=y-2 $ or $ 3y=x+8 $ .
BETA
