The normal drawn at a point `(a t1 2,-2a t_1)` of the parabola `y^2=4a x` meets it again in the point `(a t2 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)`

The normal drawn at a point (at_(1)^(2),-2at_(1)) of the parabola y^(2)=4ax meets it again in the point (at_(2)^(2),2at_(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))

The normal at the point (b t_(1)^(2), 2b t_(1)) on a parabola meets the parabola again in the point (b t_(2)^(2), 2b t_(2)) then

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 .

If the normals drawn at the points t_(1) and t_(2) on the parabola meet the parabola again at its point t_(3) , then t_(1)t_(2) equals.

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 the point " t_(1)(at_(1)^(2),2at_(1)) " on " y^(2)=4ax " meets the parabola again at the point " t_(2) " ,then " t_(1)t_(2) =

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