First slide

Tangents and normals

Question

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

Moderate

A

$(\pm 4 / \sqrt{3} - 2)$
B

$(\pm \sqrt{11 / 3}, 1)$
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.
$\begin{array}{l} y^{2} - 4 = 0 \Rightarrow y = \pm 2 \\ When y = 2, 8 + 3 x^{2} = 12 (2) \Rightarrow 3 x^{2} = 16 \Rightarrow x = \pm \frac{4}{\sqrt{3}} \end{array}$
$when y = - 2, - 8 + 3 x^{2} = - 24 \Rightarrow 3 x^{2} = - 16 . This is not possible$
$Thus, the required points are (\pm 4 / \sqrt{3}, 2)$

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