The area of the triangle formed by the tangen | vedclass

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

Question

(C) Let the point $P$ be $(at, \frac{a}{t})$.The equation of the curve is $xy = a^2$.Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} =

Vedclass · Accepted Answer

$2a^2$ sq. units

Vedclass · Answer

(C) Let the point $P$ be $(at, \frac{a}{t})$.The equation of the curve is $xy = a^2$.Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At point $P(at, \frac{a}{t})$,the slope of the tangent is $m = -\frac{a/t}{at} = -\frac{1}{t^2}$.The equation of the tangent at $P$ is $y - \frac{a}{t} = -\frac{1}{t^2}(x - at)$.$t^2y - at = -x + at$,which simplifies to $x + t^2y = 2at$.To find the $x$-intercept,set $y = 0$: $x = 2at$. So,$A = (2at, 0)$.To find the $y$-intercept,set $x = 0$: $t^2y = 2at$,so $y = \frac{2a}{t}$. So,$B = (0, \frac{2a}{t})$.The area of the right-angled triangle $	riangle ABO$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2at) 	imes (\frac{2a}{t}) = 2a^2$ sq. units.

The area of the triangle formed by the tangent to the curve $xy = a^2$ at $(x_1, y_1)$ on it and the coordinate axes is:

Similar Questions

The equation of the normal to the curve $y(1+x^{2})=2-x$ at the point where the tangent crosses the $x$-axis is

If the curves $y = \frac{\ln x}{x}$ and $y = \lambda x^2$ (where $\lambda$ is a constant) touch each other,then $\lambda$ is

The normal to the curve $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$ at any $\theta$ is such that:

The equation of the tangent to the curve $y = x + \frac{4}{x^2}$ which is parallel to the $X$-axis is:

If the subnormal at any point on the curve $y^n = ax$ is constant,then the value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module