The equation of the tangent to the curve x = | vedclass

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

Question

(D) Given the parametric equations $x = \frac{t - 1}{t + 1}$ and $y = \frac{t + 1}{t - 1}$.At $t = 2$,$x = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ and $y =

Vedclass · Accepted Answer

$9x + y - 6 = 0$

Vedclass · Answer

(D) Given the parametric equations $x = \frac{t - 1}{t + 1}$ and $y = \frac{t + 1}{t - 1}$.At $t = 2$,$x = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ and $y = \frac{2 + 1}{2 - 1} = 3$.Observe that $xy = \left(\frac{t - 1}{t + 1}ight) 	imes \left(\frac{t + 1}{t - 1}ight) = 1$.Thus,$y = \frac{1}{x}$,which implies $\frac{dy}{dx} = -\frac{1}{x^2}$.At $x = \frac{1}{3}$,the slope $m = \frac{dy}{dx} = -\frac{1}{(1/3)^2} = -9$.The equation of the tangent at $(\frac{1}{3}, 3)$ is given by $y - y_1 = m(x - x_1)$.$y - 3 = -9(x - \frac{1}{3})$.$y - 3 = -9x + 3$.$9x + y - 6 = 0$.

The equation of the tangent to the curve $x = \frac{t - 1}{t + 1}, y = \frac{t + 1}{t - 1}$ at $t = 2$ is:

Similar Questions

If $x = a \cos \theta$ and $y = b \sin \theta$,then find the value of $\left[\frac{d^2 y}{d x^2}\right]_{\theta = \frac{\pi}{4}}$.

If $x = a(t - \frac{1}{t})$ and $y = a(t + \frac{1}{t})$,then $\frac{dy}{dx} = $

If $x=e^{t}(\sin t-\cos t)$ and $y=e^{t}(\sin t+\cos t)$,then $\frac{dy}{dx}$ at $t=\frac{\pi}{3}$ is

If $x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$,where $\theta$ is a parameter,then $\frac{dy}{dx}$ at $(1,1)$ is equal to

If $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$,then $t^6\left(\frac{d u}{d t}\right)^2 \div \left(e^{2 t}(t-2)^2\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module