Question

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

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

Answer 1

The correct option is D $(\pm \frac{4}{\sqrt{3}},2)$

$\frac{dy}{dx}=\frac{2x}{4-y^2};$ For vertical tangents, $\frac{dy}{dx}\Rightarrow \infty$
$\therefore y=\pm 2$
For $y=2,x=\pm \frac{4}{\sqrt{3}}\Rightarrow$ For $y=-2,x^2=-\text{ve}$ (not possible)