Let `A(x_(1),y_(1))` and `B(x_(2),y_(2))` be two points on the parabola `y^(2) = 4ax`. If the circle with chord AB as a dimater touches the parabola, then `|y_(1)-y_(2)|` is equal to

Options

A. `4a`
B. `8a`
C. `6sqrt(2)a`
D. not a constant

Answer

B

Explanation

Equation of circle with AB as diameter is `(x-x_(1)) (x-x_(2)) + (y-y_(1)) (y-y_(2)) =0`.
Solving it with `y^(2) =4ax`, we get
`16a^(2) (y-y_(1)) (y-y_(2)) + (y^(2) -y_(1)^(2)) (y^(2)-y_(2)^(2)) =0`
`rArr (y-y_(1)) (y-y_(2)) [16a^(2) + (y+y_(1)) (y+y_(2))] = 0`
`rArr (y+y_(1)) (y+y_(2)) + 16a^(2) =0`
`rArr y^(2) + (y_(1)+y_(2)) y + y_(1)y_(2) + 16a^(2) =0`
The roots of the equation are equal if
`(y_(1)+y_(2))^(2) = 4y_(1)y_(2) + 64a^(2) rArr |y_(1)-y_(2)| = 8a`.