For the curve y^2 = 4ax,the lengths of the 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

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

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 $m = \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.$1$. Length of Subtangent = $|\frac{y}{m}| = |\frac{2at}{1/t}| = 2at^2$.$2$. Length of Tangent = $|y \sqrt{1 + \frac{1}{m^2}}| = |2at \sqrt{1 + t^2}| = 2at\sqrt{t^2 + 1}$.$3$. Length of Subnormal = $|y \cdot m| = |2at \cdot \frac{1}{t}| = 2a$.$4$. Length of Normal = $|y \sqrt{1 + m^2}| = |2at \sqrt{1 + \frac{1}{t^2}}| = |2at \frac{\sqrt{t^2 + 1}}{t}| = 2a\sqrt{t^2 + 1}$.Thus,the lengths are $2at\sqrt{t^2 + 1}, 2at^2, 2a\sqrt{t^2 + 1}, 2a$.

For the curve $y^2 = 4ax$,the lengths of the tangent,subtangent,normal,and subnormal at the point $(at^2, 2at)$ are respectively:

Similar Questions

What do the equations $x = \frac{t}{4}$ and $y = \frac{t^2}{4}$ represent?

Let $P : y^{2} = 4ax, a > 0$ be a parabola with focus $S$. Let the tangents to the parabola $P$ that make an angle of $\frac{\pi}{4}$ with the line $y = 3x + 5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A, B$ and $S$ are collinear is:

For the parabola represented in the parametric form by $x=t^2+t+1$ and $y=t^2-t+1$,the length of the latus rectum is

What is the equation of the normal to the curve $y^2 = 4ax$ at the point $(a, 2a)$?

If the tangents and normals at the extremities of a focal chord of a parabola $y^2 = 4ax$ intersect at $(x_1, y_1)$ and $(x_2, y_2)$ respectively,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module