Normals `A O ,AA_1a n dAA_2` are drawn to the parabola `y^2=8x` from the point `A(h ,0)` . If triangle `O A_1A_2` is equilateral then the possible value of `h` is 26 (b) 24 (c) 28 (d) none of these

Options

A. 26
B. 24
C. 28
D. none of these

Answer

C

Explanation

(3) Let `A_(1)-=(2at_(1)^(2),4t_(1))andA_(2)-=(2t_(1)^(2),-4t_(1))`.
Clearly, `angleA_(1)OA=pi//6`. Therefore,
`(2)/(t_(1))=(1)/(sqrt(3))`
The equation of normal of normal at `A_(1)` is
`y=-t_(1)x+4t_(1)+2t_(1)^(3)`
`orh=4+2t_(1)^(2)=4+2(12)`
=28