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 = \pm2 $
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(\pm4 /\sqrt{3} , 2\right)$