Application of Derivatives 1 Question 11
####11. The point(s) on the curve $y^{3}+3 x^{2}=12 y$, where the tangent is vertical, is (are)
(2002, 2M)
(a) $\pm \frac{4}{\sqrt{3}},-2$
(b) $\pm \sqrt{\frac{11}{3}}, 0$
(c) $(0,0)$
(d) $\pm \frac{4}{\sqrt{3}}, 2$
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Answer:
Correct Answer: 11. $1: 16$
Solution:
- Given, $y^{3}+3 x^{2}=12 y$
On differentiating w.r.t. $x$, we get
$ \begin{array}{rlrl} \Rightarrow & & 3 y^{2} \frac{d y}{d x}+6 x & =12 \frac{d y}{d x} \\ \Rightarrow & \frac{d y}{d x} & =\frac{6 x}{12-3 y^{2}} \end{array} $
$\Rightarrow \quad \frac{d x}{d y}=\frac{12-3 y^{2}}{6 x}$
For vertical tangent, $\frac{d x}{d y}=0$
$\Rightarrow \quad 12-3 y^{2}=0 \quad \Rightarrow \quad y= \pm 2$
On putting, $y=2$ in Eq. (i), we get $x= \pm \frac{4}{\sqrt{3}}$ and again putting $y=-2$ in Eq. (i), we get $3 x^{2}=-16$, no real solution.
So, the required point is $\pm \frac{4}{\sqrt{3}}, 2$.
