Show that the normal at a point `(at^2_1, 2at_1)` on the parabola `y^2 = 4ax` cuts the curve again at the point whose parameter `t_2 = -t_1 - 2/t_1`.

Show that the normal at the point (3t,(4)/(t)) to the curve xy=12 cuts the curve again at the point whose parameter t_(1) is given by t_(1)=-(16)/(9t^(3))

if the normal at the point t_(1) on the parabola y^(2) = 4ax meets the parabola again in the point t_(2) then prove that t_(2) = - ( t_(1) + 2/t_(1))

The normal at t_(1) and t_(2) on the parabola y^(2)=4ax intersect on the curve then t_(1)t_(2)

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

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

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

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 drawn at a point (a t_1^2,-2a t_1) of the parabola y^2=4a x meets it again in the point (a t_2^2,2a t_2), then t_2=t_1+2/(t_1) (b) t_2=t_1-2/(t_1) t_2=-t_1+2/(t_1) (d) t_2=-t_1-2/(t_1)