The length of the focal chord of the parabola | vedclass

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

Question

(C) Let the focal chord make an angle $	heta$ with the $x$-axis. The length of the focal chord is given by $L = 4a \csc^2 	heta$.The distance $p$ of

Vedclass · Accepted Answer

$\frac{4a^3}{p^2}$

Vedclass · Answer

(C) Let the focal chord make an angle $	heta$ with the $x$-axis. The length of the focal chord is given by $L = 4a \csc^2 	heta$.The distance $p$ of the focal chord from the vertex $(0,0)$ is the perpendicular distance from the origin to the line passing through the focus $(a,0)$ with inclination $	heta$.The equation of the focal chord is $(x - a) \sin 	heta - y \cos 	heta = 0$.The distance $p$ from the origin $(0,0)$ is $p = |\frac{-a \sin 	heta}{\sqrt{\sin^2 	heta + \cos^2 	heta}}| = |a \sin 	heta|$.Thus,$\sin 	heta = \frac{p}{a}$,which implies $\csc 	heta = \frac{a}{p}$.Substituting this into the length formula: $L = 4a (\frac{a}{p})^2 = \frac{4a^3}{p^2}$.

The length of the focal chord of the parabola $y^2 = 4ax$ at a distance $p$ from the vertex is:

Similar Questions

If the point $(a, 2a)$ is an interior point of the region bounded by the parabola $y^2 = 16x$ and the double ordinate through the focus,then

If one end of a focal chord of the parabola $y^2 = \frac{8}{a} x$ $(a > 0)$ is at $(1, 4)$,then the length of this focal chord is

The length of the subnormal to the parabola $y^2 = 4ax$ at any point is equal to

If $mx - y + c = 0$ is a normal at a point $P$ on the parabola $y^2 = 16x$ and the focal distance of $P$ is $40$ units,then $|c| =$

Find the equation of the normal to the curve $x^{2}=4y$ which passes through the point $(1, 2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module