Prove that the feet of the normals drawn from the point `(h, k)` to the parabola `y^2 -4ax` lie on the curve `xy-(h-2a)y-2ak=0`.

Three normals drawn from a point (h k) to parabola y^2 = 4ax

The number of normals drawn from the point (6, -8) to the parabola y^2 - 12y - 4x + 4 = 0 is

The feet of the normals to y^(2)= 4ax from the point (6a,0) are

If the tangents drawn from the point (0, 2) to the parabola y^2 = 4ax are inclined at angle (3pi)/4 , then the value of 'a' is

Three normals are drawn from the point (7, 14) to the parabola x^2-8x-16 y=0 . Find the coordinates of the feet of the normals.

Show that the locus of points such that two of the three normals drawn from them to the parabola y^2 = 4ax coincide is 27ay^2 = 4(x-2a)^3 .

If two of the three feet of normals drawn from a point to the parabola y^2=4x are (1, 2) and (1,-2), then find the third foot.

If two of the three feet of normals drawn from a point to the parabola y^2=4x are (1, 2) and (1,-2), then find the third foot.

If t_(1),t_(2),t_(3) are the feet of normals drawn from (x_(1),y_(1)) to the parabola y^(2)=4ax then the value of t_(1)t_(2)t_(3) =

P & Q are the points of contact of the tangents drawn from the point T to the parabola y^(2) = 4ax . If PQ be the normal to the parabola at P, prove that TP is bisected by the directrix.