1.

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

Answer»

`4a`
`8A`
`6sqrt(2)a`
not a constant

Solution :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`.


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