The length of the latus-rectum of the parabol | vedclass

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(D) The focus $S$ is $\left( \frac{u^2}{2g} \sin 2\alpha, -\frac{u^2}{2g} \cos 2\alpha \right)$ and the directrix is $y = \frac{u^2}{2g}$.The distance

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$\frac{2u^2}{g} \cos^2 \alpha$

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(D) The focus $S$ is $\left( \frac{u^2}{2g} \sin 2\alpha, -\frac{u^2}{2g} \cos 2\alpha ight)$ and the directrix is $y = \frac{u^2}{2g}$.The distance $d$ from the focus to the directrix is the perpendicular distance from the point $S$ to the line $y = \frac{u^2}{2g}$.$d = \left| \frac{u^2}{2g} - \left( -\frac{u^2}{2g} \cos 2\alpha ight) ight| = \frac{u^2}{2g} (1 + \cos 2\alpha)$.Using the trigonometric identity $1 + \cos 2\alpha = 2 \cos^2 \alpha$,we get $d = \frac{u^2}{2g} (2 \cos^2 \alpha) = \frac{u^2}{g} \cos^2 \alpha$.The length of the latus-rectum of a parabola is $2 	imes (	ext{distance from focus to directrix}) = 2d$.Length $= 2 	imes \frac{u^2}{g} \cos^2 \alpha = \frac{2u^2}{g} \cos^2 \alpha$.

The length of the latus-rectum of the parabola whose focus is $\left( \frac{u^2}{2g} \sin 2\alpha, -\frac{u^2}{2g} \cos 2\alpha \right)$ and directrix is $y = \frac{u^2}{2g}$,is

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