Question

The point(s) on the curve $y^3+3x^2=12y$ where the tangent is vertical, is (are)

• A. $\left(\pm \frac{4}{\sqrt{3}},-2\right)$
• B. $\left(\pm \sqrt{\frac{11}{3}},1\right)$
• C. (0, 0)
• D. $\left(\pm \frac{4}{\sqrt{3}},2\right)$

Answer 1

Answer: (D)

The given curve is $y^3+3x^2=12y$
$\Rightarrow 3y^2\frac{dy}{dx}+6x=12\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{2x}{4-y^2}$
$\frac{dy}{dx}=\frac{1}{0}\Rightarrow 4-y^2=0$
$\Rightarrow y=\pm 2$
For $y=2,x^2=\frac{24-8}{3}=\frac{16}{3}$
$\Rightarrow x=\pm 4\sqrt{3}$
For $y=-2,x^2=\frac{-24+8}{3}=-ve$ (not possible)
$\therefore$ Req. points are $\left(\pm 4/\sqrt{3},2\right)$