If x and y are connected parametrically by th | vedclass

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Question

(A) Given equations are $x = a \sec 	heta$ and $y = b 	an 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = a \sec 	heta

Vedclass · Accepted Answer

$\frac{b}{a} \csc 	heta$

Vedclass · Answer

(A) Given equations are $x = a \sec 	heta$ and $y = b 	an 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = a \sec 	heta 	an 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = b \sec^2 	heta$. Using the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{b \sec^2 	heta}{a \sec 	heta 	an 	heta}$. Simplifying the expression: $\frac{dy}{dx} = \frac{b}{a} \cdot \frac{\sec 	heta}{	an 	heta} = \frac{b}{a} \cdot \frac{1/\cos 	heta}{\sin 	heta / \cos 	heta} = \frac{b}{a} \cdot \frac{1}{\sin 	heta} = \frac{b}{a} \csc 	heta$.

List-$I$	List-$II$
$(i)$ $x = a(\theta - \sin \theta), y = a(1 - \cos \theta)$	$(A)$ $4\sqrt{3}$
(ii) $x = 3\cos \theta - 2\cos^3 \theta, y = 3\sin \theta - 2\sin^3 \theta$	$(B)$ $-\frac{1}{3\sqrt{3}}$
(iii) $x = 3\cos \theta - \cos^3 \theta, y = 3\sin \theta - \sin^3 \theta$	$(C)$ $\sqrt{3}$
(iv) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \sec \theta$ and $y = b \tan \theta$.

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