Question

'$O$' is the vertex of the parabola $y^2=8x$ and L is the upper end of the latus rectum. If LH is drawn perpendicular to $OL$ meeting $OX$ in $H$, then the length of the double ordinate through $H$ is $\lambda \sqrt{5}$ where $\lambda$ is equal to

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is A $2$

Equation of LH
$y-2a=\frac{-1}{2}(x-a)$
put $y=0$
$x=5a$
$\therefore y^2=4ax=\left(4a\right)\left(5a\right)=20a^2$
$y=2\sqrt{5}a$
$2y=4\sqrt{5}$
Slope of OL$=2$
Slope of $LH=\frac{-1}{2}$