#### Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 5

$\frac{dy}{dx}=-\frac{a}{b}$

Hint:

Use chain rule

Given:

$x=b \sin ^{2} \theta \\$ , $y=a \cos ^{2} \theta \\$

Solution:

$x=b \sin ^{2} \theta \\$

$\frac{d x}{d \theta}=b \frac{d\left(\sin ^{2} \theta\right)}{d \theta} \\$

$\frac{d x}{d \theta}=b \times\left[\frac{d \sin ^{2} \theta}{d \sin \theta} \times \frac{d \sin \theta}{d \theta}\right] \$                                                                            [Using chain rule]

$\ =b \times[2 \sin \theta \times \cos \theta] \\$

\begin{aligned} &&\frac{d x}{d \theta}=2 b \sin \theta \cos \theta \end{aligned}

$y=a \cos ^{2} \theta \\$

$\frac{d y}{d \theta}=a \cdot \frac{d\left(\cos ^{2} \theta\right)}{d \theta}$

\begin{aligned} &\\ &=a \cdot \frac{d \cos ^{2} \theta}{d \cos \theta} \times \frac{d \cos \theta}{d \theta} \end{aligned}                                                                        [Using chain rule]

$=a(2 \cos \theta)(-\sin \theta) \\$

$\frac{d y}{d \theta}=-2 a \sin \theta \cos \theta$

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