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  • Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 10 maths textbook solution

Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 10 maths textbook solution

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Answer:  Point P \left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)

Hint: Find  \frac{dy}{dx} and then use this formula for slope of tangent.

Given:    \mathrm{x}=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \quad 0<\theta<\frac{\pi}{2} .  

                The tangent to C is parallel to the chord joining the points (a, 0) and (0, a).

Solution:

\begin{aligned} &x=a \cos ^{3} \theta \\ &\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta \\ &y=a \sin ^{3} \theta \\ &\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta \end{aligned}

Now, \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}

\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{3 a \sin ^{2} \theta \cdot \cos \theta}{-3 a \cos ^{2} \theta \cdot \sin \theta} \\ &\Rightarrow \frac{d y}{d x}=-\tan \theta \end{aligned}

Now, slope of chord =\frac{a-0}{0-a}

Slope of chord =\frac{a}{-a}=-1 

Slope of chord =\frac{d y}{d x}

\begin{aligned} &\Rightarrow-1=-\tan \theta \\ &\Rightarrow \tan \theta=1 \\ &\Rightarrow \tan \theta=\tan \frac{\pi}{4} \\ &\Rightarrow \theta=\frac{\pi}{4} \end{aligned}

Now, x=a \cos ^{3} \theta

\begin{aligned} &=a \cos ^{3} \frac{\pi}{4} \\ &=a\left(\frac{1}{\sqrt{2}}\right)^{2} \\ &=\frac{a}{2 \sqrt{2}} \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{aligned}

Now, y=a \sin ^{3} \theta

\begin{aligned} &=a \sin ^{3} \frac{\pi}{4} \\ &=a\left(\frac{1}{\sqrt{2}}\right)^{2} \\ &=\frac{a}{2 \sqrt{2}} \quad\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{aligned}

Hence Point P is \left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)

Posted by

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