If x = a(t - frac{1}{t}) and y = a(t + frac{1 | vedclass

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

Question

(C) Given equations are $x = a(t - \frac{1}{t})$ $(i)$ and $y = a(t + \frac{1}{t})$ $(ii)$.Squaring both equations,we get $x^2 = a^2(t^2 - 2 + \frac{1

Vedclass · Accepted Answer

$\frac{x}{y}$

Vedclass · Answer

(C) Given equations are $x = a(t - \frac{1}{t})$ $(i)$ and $y = a(t + \frac{1}{t})$ $(ii)$.Squaring both equations,we get $x^2 = a^2(t^2 - 2 + \frac{1}{t^2})$ and $y^2 = a^2(t^2 + 2 + \frac{1}{t^2})$.Subtracting $(i)^2$ from $(ii)^2$,we get $y^2 - x^2 = a^2(t^2 + 2 + \frac{1}{t^2} - (t^2 - 2 + \frac{1}{t^2})) = a^2(4) = 4a^2$.Differentiating $y^2 - x^2 = 4a^2$ with respect to $x$,we get $2y \frac{dy}{dx} - 2x = 0$.Therefore,$2y \frac{dy}{dx} = 2x$,which simplifies to $\frac{dy}{dx} = \frac{x}{y}$.

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

Similar Questions

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$,then $\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=$

If $x=f(t)$ and $y=g(t)$ are differentiable functions of $t$,then $\frac{d^2 y}{d x^2}$ is

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

The slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module