Question

The points on the curve y³ + 3x² = 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)

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

Answer 2

y³ + 3x² = 12y

$\frac{dy}{dx} = \frac{6x}{3(4 - y^{2})}$. For vertical tangents y = ± 2

But when y = – 2 x² = –ve, it is not possible

for y = 2, x² = $\frac{16}{3}$. Hence point are $\left( \pm \frac{4}{\sqrt{3}},2 \right)$