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

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 $\left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)$