If normal at point `P` on the parabola `y^2=4a x ,(a >0),` meets it again at `Q` in such a way that `O Q` is of minimum length, where `O` is the vertex of parabola, then ` O P Q` is

The vertex of the parabola y^2 + 4x = 0 is

A is a point on the parabola y^2=4a x . The normal at A cuts the parabola again at point Bdot If A B subtends a right angle at the vertex of the parabola, find the slope of A Bdot

A normal drawn to the parabola y^2=4a x meets the curve again at Q such that the angle subtended by P Q at the vertex is 90^0dot Then the coordinates of P can be (a) (8a ,4sqrt(2)a) (b) (8a ,4a) (c) (2a ,-2sqrt(2)a) (d) (2a ,2sqrt(2)a)

If the normals to the parabola y^2=4a x at P meets the curve again at Q and if P Q and the normal at Q make angle alpha and beta , respectively, with the x-axis, then t a nalpha(tanalpha+tanbeta) has the value equal to 0 (b) -2 (c) -1/2 (d) -1

If the tangents at the points Pa n dQ on the parabola y^2=4a x meet at T ,a n dS is its focus, the prove that S P ,S T ,a n dS Q are in GP.

If normal to parabola y^(2)=4ax at point P(at^(2),2at) intersects the parabola again at Q, such that sum of ordinates of the points P and Q is 3, then find the length of latus ectum in terms of t.

If the normals at P, Q, R of the parabola y^2=4ax meet in O and S be its focus, then prove that .SP . SQ . SR = a . (SO)^2 .

The normal at the point P(ap^2, 2ap) meets the parabola y^2= 4ax again at Q(aq^2, 2aq) such that the lines joining the origin to P and Q are at right angle. Then (A) p^2=2 (B) q^2=2 (C) p=2q (D) q=2p

The tangent at any point P onthe parabola y^2=4a x intersects the y-axis at Qdot Then tangent to the circumcircle of triangle P Q S(S is the focus) at Q is

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))