If a focal chord of the parabola y^2 = 4ax ma | vedclass

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(C) The equation of a focal chord of the parabola $y^2 = 4ax$ passing through the focus $(a, 0)$ with slope $m = 	an 	heta$ is given by:$y - 0 = 	a

Vedclass · Accepted Answer

$a \sin 	heta$

Vedclass · Answer

(C) The equation of a focal chord of the parabola $y^2 = 4ax$ passing through the focus $(a, 0)$ with slope $m = 	an 	heta$ is given by:$y - 0 = 	an 	heta (x - a)$$y = 	an 	heta (x - a)$$	an 	heta \cdot x - y - a 	an 	heta = 0$The length of the perpendicular from the vertex $(0, 0)$ to this line is given by the formula $d = \left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} ight|$.Substituting the values:$d = \left| \frac{	an 	heta (0) - 1(0) - a 	an 	heta}{\sqrt{	an^2 	heta + (-1)^2}} ight|$$d = \left| \frac{-a 	an 	heta}{\sqrt{\sec^2 	heta}} ight|$$d = \left| \frac{-a 	an 	heta}{\sec 	heta} ight|$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$ and $\sec 	heta = \frac{1}{\cos 	heta}$,we get:$d = a \sin 	heta$

If a focal chord of the parabola $y^2 = 4ax$ makes an angle $\theta$ with its axis,then the length of the perpendicular from the vertex to this chord is......

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