SATHEE: Application of Derivatives 1 Question 11

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) $\pm \sqrt{\frac{11}{3}}, 0$

(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

$\begin{aligned} \Rightarrow & 3 y^{2} \frac{d y}{d x} + 6 x & = 12 \frac{d y}{d x} \\ \Rightarrow & \frac{d y}{d x} & = \frac{6 x}{12 - 3 y^{2}} \end{aligned}$

$\Rightarrow \frac{d x}{d y} = \frac{12 - 3 y^{2}}{6 x}$

For vertical tangent, $\frac{d x}{d y} = 0$

$\Rightarrow 12 - 3 y^{2} = 0 \Rightarrow y = \pm 2$

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

Application of Derivatives 1 Question 11

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Application of Derivatives 1 Question 11

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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