In the parabola `y^2 = 4x`, the ends of the double ordinate through the focus are P and Q. Let O be the vertex. Then the length of the double ordinate PQ is

'O' is the vertex of the parabola y^(2)=4ax&L is the upper end of the latus rectum.If LH is drawn perpendicular to OL meeting OX in H, prove that the length of the double ordinate through H is 4a sqrt(5).

Ends of the double ordinate of y=16x which is at a distance of 8 units from the vertex is

‘O’ is the vertex of the parabola y^(2) = 4ax & L is the upper end of the latus rectum. If LH is drawn perpendicular to OL meeting OX in H, prove that the length of the double ordinate through H is 4asqrt5 .

Let S is the focus of the parabola y^(2) = 4ax and X the foot of the directrix, PP’ is a double ordinate of the curve and PX meets the curve again in Q. Prove that P’Q passes through focus.

Let S is the focus of the parabola y^(2)=4ax and X the foot of the directrix,PP' is a double ordinate of the curve and PX meets the curve again in Q. Prove that P'Q passes through focus.

Consider a parabola y^(2)=4x with vertex at A and focus at S. PQ is a chord of the parabola which is normal at point P. If the abscissa and the ordinate of the point P are equal, then the square of the length of the diameter of the circumcircle of triangle PSQ is

For the parabola y^(2)=4px find the extremities of a double ordinate of length 8p. Prove that the lines from the vertex to its extremities are at right angles.