If x=frac{1-t^2}{1+t^2} and y=frac{2at}{1+t^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.Substitute $t=	an 	heta$,then $x=\cos 2	heta$ and $y=a \sin 2	heta$.Differentiating wi

Vedclass · Accepted Answer

$\frac{a(t^2-1)}{2t}$

Vedclass · Answer

(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.Substitute $t=	an 	heta$,then $x=\cos 2	heta$ and $y=a \sin 2	heta$.Differentiating with respect to $	heta$:$\frac{dx}{d	heta} = -2 \sin 2	heta$ and $\frac{dy}{d	heta} = 2a \cos 2	heta$.Using the chain rule,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2a \cos 2	heta}{-2 \sin 2	heta} = -a \cot 2	heta$.Since $\cot 2	heta = \frac{1}{	an 2	heta} = \frac{1-	an^2 	heta}{2 	an 	heta} = \frac{1-t^2}{2t}$,$\frac{dy}{dx} = -a \left( \frac{1-t^2}{2t} ight) = \frac{a(t^2-1)}{2t}$.

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$,then $\frac{dy}{dx}=$

Similar Questions

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

If $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$ where $\theta \in [0, 2 \pi]$,then find the value of $\frac{d^{2} y}{dx^{2}}$ at $\theta = \pi$.

If $x = \sin t \cos 2t$ and $y = \cos t \sin 2t$,then at $t = \frac{\pi}{4}$,the value of $\frac{dy}{dx}$ is equal to

If $\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$,then the value of $\frac{dy}{dx}$ is

The equation of the normal to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module