JEE Advance - Mathematics (2002 - No. 1)
The point(s) in the curve $${y^3} + 3{x^2} = 12y$$ where the tangent is vertical, is (are)
$$\left( { \pm {4 \over {\sqrt 3 }}, - 2} \right)$$
$$\left( { \pm \sqrt {{{11} \over 3}} ,1} \right)$$
$$(0,0)$$
$$\left( { \pm {4 \over {\sqrt 3 }}, 2} \right)$$
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