A focal chord of parabola y2 = 4x is inclined at an angle of  with the +ive direction of x-axis, then the slope of normal drawn at the ends of focal chord will satisfy the equation


Correct option is

m2 + 2m – 1 = 0

Let A, B be the points (t12, 2t1) and (t22, 2t2), (a = 1) be two pints on the parabola y2 = 4x.

Since AB is a focal chord, t­1 t­2 = –1.


Also slope of chord y(t1 + t2) – 2x – 2at­1t2 = 0 is

Hence t1, t2 are the roots of

          m2 – 2m – 1 = 0

Slopes of normal’s at A and B are – t1, – t2 which are roots of

      (–m)2 – 2(–m) – 1 = 0

          m2 + 2m – 1 = 0



A tangent and a normal are drawn at the point P(16, 16) of the parabola y2 = 16x which cut the axis of the parabola at the points A and B respectively. If the center of the circle through P, A and B is C, then angle between PC and axis of x is


If x + y = k is normal to y2 = 12x, then k is


A circle drawn on any focal chord AB of the parabola y2 = 4axas diameter cuts the parabola again at and D. If the parameters of the points A, B, C, D be t1, t2 t3 and t1 respectively, then the value of t3 t4 is


The length of normal chord which subtends an angle of 900 at the vertex of the parabola y2 = 4x is


If two different tangents of y2 = 4x are the normal’s to the parabola x2 = 4ay, then


Find the locus of the mid-points of the chord of the parabola y2 = 4ax which subtend a right angle at the vertex.


If the parabola C and D intersect at a point A on the line L1, then equation of the tangent point L at A to the parabola D is


If a > 0, the angle subtended by the chord AB at the vertex of the parabola is


P is a point on the circle C, the perpendicular PQ to the major axis of the ellipse E meets the ellipse at M, then  is equal to


Equation of the diameter of the ellipse conjugate to the diameter respected by L is