Applications Of Derivatives Question 139
Question: At what point on the curve $ x^{3}-8a^{2}y=0 $ , the slope of the normal is $ \frac{-2}{3} $
[RPET 2002]
Options:
A) $ (a,,a) $
B) $ (2a,,-a) $
C) $ (2a,,a) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ x^{3}-8a^{2}y=0 $
therefore $ 3x^{2}-8a^{2},.,\frac{dy}{dx}=0 $
therefore $ 3x^{2}=8a^{2},.,\frac{dy}{dx} $
therefore $ \frac{dy}{dx}=\frac{3x^{2}}{8a^{2}} $
Slope of the normal = $ -\frac{1}{( \frac{dy}{dx} )} $ = $ -\frac{1}{\frac{3x^{2}}{8a^{2}}} $
$ =-\frac{8a^{2}}{3x^{2}} $
Given $ \frac{-8a^{2}}{3x^{2}}=\frac{-2}{3} $ \ $ (x,,y)=(2a,,a) $ .
BETA
