Applications Of Derivatives Question 228
Question: The point(s) on the curve $ y^{3}+3x^{2}=12y $ where the tangent is vertical (parallel to y-axis), is (are)
[IIT Screening 2002]
Options:
A) $ ( \pm \frac{4}{\sqrt{3}},-2 ) $
B) $ ( \pm \frac{\sqrt{11}}{3},1 ) $
C) $ (0,,0) $
D) $ ( \pm \frac{4}{\sqrt{3}},2 ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ y^{3}+3x^{2}=12y $
therefore $ 3y^{2},.,\frac{dy}{dx}+6x=12,.,\frac{dy}{dx} $
therefore $ \frac{dy}{dx}(3y^{2}-12)+6x=0 $
therefore $ \frac{dy}{dx}=\frac{6x}{12-3y^{2}} $
therefore $ \frac{dx}{dy}=\frac{12-3y^{2}}{6x} $
Since tangent is parallel to y-axis
$ \therefore $ $ \frac{dx}{dy}=0 $
therefore $ 12-3y^{2}=0 $ or $ y=\pm 2 $ .
Then $ x=\pm \frac{4}{\sqrt{3}} $ . At $ ( \pm \frac{4}{\sqrt{3}},,-2 ); $ the equation of curve doesn-t satisfy.
BETA
