Question

The point (s) on the curve $y^{3} + 3x^{2} = 12y$ where the tangent is vertical (parallel to y-axis), is are

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

Answer 1

Answer: (D)

$\left( \pm \frac{4}{\sqrt{3}},2 \right)$

Answer 2

$y^{3} + 3x^{2} = 12y$

⇒ $3y^{2}.\frac{dy}{dx} + 6x = 12.\frac{dy}{dx}$ ⇒ $\frac{dy}{dx}(3y^{2} - 12) + 6x = 0$

⇒ $\frac{dy}{dx} = \frac{6x}{12 - 3y^{2}}$ ⇒ $\frac{dx}{dy} = \frac{12 - 3y^{2}}{6x}$

Tangent is parallel to y-axis, $\frac{dx}{dy} = 0$ ⇒ $12 - 3y^{2}$ = 0 or $y = \pm 2.$ Then $x = \pm \frac{4}{\sqrt{3}}$, for $y = 2$

$y = - 2$does not satisfy the equation of the curve,

$\therefore$ The point is $\left( \pm \frac{4}{\sqrt{3}},2 \right)$