First slide

Tangents and normals

Question

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

Easy

A

$(\pm \frac{4}{\sqrt{3}}, 2) .$
B

$(\pm \sqrt{\frac{11}{3}}, 1)$
C

$(0, 0)$
D

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

Solution

Differentiating the given curve $w . r . t . x,$ we get
$3 y^{2} \frac{d y}{d x} + 6 x = 12 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = - \frac{2 x}{y^{2} - 4}$
At point where the tangent (s) is (are) vertical, $\frac{d y}{d x}$ is not defined , i.e. at those points ,
$y^{2} - 4 = 0 \Rightarrow y = \pm 2$
When $y = - 2, - 8 + 3 x^{2} = - 24 \Rightarrow 3 x^{2} = - 16.$
This is not possible.
Thus , the required points are $(\pm \frac{4}{\sqrt{3}}, 2) .$

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