Find the equations of the tangent and normal | vedclass

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

Question

(A) The equation of the given parabola is $y^{2}=4ax$. On differentiating $y^{2}=4ax$ with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies

Vedclass · Accepted Answer

Tangent: $ty = x + at^{2}$,Normal: $y = -tx + 2at + at^{3}$

Vedclass · Answer

(A) The equation of the given parabola is $y^{2}=4ax$. On differentiating $y^{2}=4ax$ with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$. The slope of the tangent at $(at^{2}, 2at)$ is $\left. \frac{dy}{dx} \right|_{(at^{2}, 2at)} = \frac{2a}{2at} = \frac{1}{t}$. The equation of the tangent is $y - 2at = \frac{1}{t}(x - at^{2})$,which simplifies to $ty - 2at^{2} = x - at^{2}$,or $ty = x + at^{2}$. The slope of the normal is the negative reciprocal of the tangent slope,which is $-t$. The equation of the normal is $y - 2at = -t(x - at^{2})$,which simplifies to $y - 2at = -tx + at^{3}$,or $y = -tx + 2at + at^{3}$.

Find the equations of the tangent and normal to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$.

Similar Questions

The locus of the point of intersection of the perpendicular tangents to the parabola $x^2 = 4ay$ is

If $t_1$ and $t_2$ are the parameters of the end points of a focal chord for the parabola $y^2 = 4ax$, then which one is true?

Tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$. The length of the intercept made by these tangents on the line $x = 2$ is:

The parabola $y = x^2 + px + q$ cuts the straight line $y = 2x - 3$ at a point with abscissa $1$. If the distance between the vertex of the parabola and the $x$-axis is least,then:

The condition that the line $ax + by + c = 0$ is tangent to the parabola $y^2 = 4ax$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module