If the normal to the parabola `y^2=4a x` at point `t_1` cuts the parabola again at point `t_2` , then prove that `t2 2geq8.`

IF the normal to the parabola y^2=4ax at point t_1 cuts the parabola again at point t_2 , prove that t_2^2ge8

If the normal to the parabola y^(2)=4ax at point t_(1) cuts the parabola again at point t_(2) .Then the minimum value of t_(2)^(2) is

The normal to the parabola y^(2)=8ax at the point (2, 4) meets the parabola again at the point

If the normal to the parabola y^(2)=4ax at the point (at^(2),2at) cuts the parabola again at (aT^(2),2aT) then

If the normal at(1, 2) on the parabola y^(2)=4x meets the parabola again at the point (t^(2), 2t) then the value of t, is

The normal to the parabola y^(2)=4x at P (1, 2) meets the parabola again in Q, then coordinates of Q are

The normal at a point with parameter 2 on y^(2)=6x again meets the parabola at point