Question
The point(s) on the curve $${y^3} + 3{x^2} = 12y$$ 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.
$$\left( {0,0} \right)$$
D.
$$\left( { \pm \frac{4}{{\sqrt 3 }},2} \right)$$
Answer :
$$\left( { \pm \frac{4}{{\sqrt 3 }},2} \right)$$
Solution :
$$\eqalign{
& {\text{The}}\,{\text{given}}\,{\text{curve}}\,{\text{is}}\,{y^3} + 3{x^2} = 12y \cr
& \Rightarrow 3{y^2}\frac{{dy}}{{dx}} + 6x = 12\frac{{dy}}{{dx}} \Rightarrow \quad \frac{{dy}}{{dx}} = \frac{{2x}}{{4 - {y^2}}} \cr
& {\text{For}}\,{\text{vertical}}\,{\text{tangents}}\,\frac{{dy}}{{dx}} = \frac{1}{0} \Rightarrow 4 - {y^2} = 0 \Rightarrow y = \pm 2 \cr
& {\text{For}}\,y = 2,{x^2} = \frac{{24 - 8}}{3} = \frac{{16}}{3} \Rightarrow x = \pm \frac{4}{{\sqrt 3 }} \cr
& {\text{For}}\,y = - 2,{x^2} = \frac{{ - 24 + 8}}{3} = - ve\,\left( {{\text{not}}\,{\text{possible}}} \right) \cr
& \therefore {\text{Req}}{\text{.}}\,{\text{points}}\,{\text{are}}\,\left( { \pm \frac{4}{{\sqrt 3 }},2} \right) \cr} $$