Tangent at any point theta on the curve x=35 | vedclass

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Question

(B) Given,$x=35 \sec 	heta$ and $y=35 	an 	heta$. Point $P = (35 \sec 	heta, 35 	an 	heta)$. Differentiating $x$ and $y$ with respect to $	heta

Vedclass · Accepted Answer

$y \sin 	heta=x-35 \cos 	heta$

Vedclass · Answer

(B) Given,$x=35 \sec 	heta$ and $y=35 	an 	heta$. Point $P = (35 \sec 	heta, 35 	an 	heta)$. Differentiating $x$ and $y$ with respect to $	heta$: $\frac{dx}{d	heta} = 35 \sec 	heta 	an 	heta$ $\frac{dy}{d	heta} = 35 \sec^2 	heta$ Slope of the tangent $m = \frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{35 \sec^2 	heta}{35 \sec 	heta 	an 	heta} = \frac{\sec 	heta}{	an 	heta} = \frac{1}{\sin 	heta}$. Equation of the tangent at $P$ is $y - y_1 = m(x - x_1)$: $y - 35 	an 	heta = \frac{1}{\sin 	heta} (x - 35 \sec 	heta)$ $y \sin 	heta - 35 	an 	heta \sin 	heta = x - 35 \sec 	heta$ $y \sin 	heta - 35 \frac{\sin^2 	heta}{\cos 	heta} = x - \frac{35}{\cos 	heta}$ $y \sin 	heta = x + 35 \frac{\sin^2 	heta}{\cos 	heta} - \frac{35}{\cos 	heta}$ $y \sin 	heta = x + \frac{35(\sin^2 	heta - 1)}{\cos 	heta}$ $y \sin 	heta = x + \frac{35(-\cos^2 	heta)}{\cos 	heta}$ $y \sin 	heta = x - 35 \cos 	heta$. Thus,the correct option is $B$.

Tangent at any point $\theta$ on the curve $x=35 \sec \theta, y=35 \tan \theta$ is

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