Find the length of the tangent and the sub-ta | vedclass

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

Question

(A) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.At th

Vedclass · Accepted Answer

$2at\sqrt{t^2 + 1}, 2at^2$

Vedclass · Answer

(A) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.At the point $(at^2, 2at)$,the slope of the tangent is $m = \frac{2a}{2at} = \frac{1}{t}$.The length of the tangent is given by $|y| \sqrt{1 + (\frac{dx}{dy})^2} = |2at| \sqrt{1 + t^2} = 2at\sqrt{t^2 + 1}$ (assuming $t > 0$).The length of the sub-tangent is given by $|y \frac{dx}{dy}| = |2at \cdot t| = 2at^2$.

Find the length of the tangent and the sub-tangent for the curve $y^2 = 4ax$ at the point $(at^2, 2at)$.

Similar Questions

What is the equation of the tangent to the curve $x = a \cos^3 t, y = a \sin^3 t$ at point $t$?

If curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect orthogonally,then $a^2$ equals:

If the tangent to the curve $xy + ax + by = 0$ at $(1, 1)$ makes an angle $\tan^{-1}(2)$ with the $x$-axis,then $\left( \frac{a + b}{ab} \right)$ is equal to:

The equation of the normal to the curve $y=x \log x$ parallel to $2x-2y+3=0$ is

The equation of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at the point $(x_1, y_1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module