question_answer

'O' is the vertex of the parabola \[{{y}^{2}}\text{ }=\text{ }8x\] 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

Correct Answer: D

Solution :

 Equation of LH \[(y-2a)\,=\frac{-1}{2}\,(x-a)\] Put \[y=0\] \[\Rightarrow \,x=5a\] \[\therefore \,\,{{y}^{2}}=4ax\,(4a)(5a)\,=20{{a}^{2}}\] \[y=2\sqrt{5}a\] \[\Rightarrow \,\,2y=4\sqrt{5}\,a\] [Slope of OL = 2; Slope of LH \[=\frac{-1}{2}\]]