Let PQ be a focal chord of the parabola x^2 = 16y of length 25 , making an acute angle with the positive x -axis. Let the abscissa of P be positive and M be the point on the line segment PQ such that PM : MQ = 1 : 4 . The y -coordinate of the point where the line passing through M and perpendicular to the line PQ intersects the y -axis is

Step-by-step solution

The equation of the parabola is x^2 = 16y , which is of the form x^2 = 4ay with a = 4 . The parametric coordinates of the endpoints of a focal chord are P(2at, at^2) and Q (- 2a t , a t^2 ) . The length of the focal chord is given by a (t + 1 t )^2 . Given that the length is 25 , we have: 4 (t + 1 t )^2 = 25 t + 1 t = 5 2 (since t + 1 t > 0 for positive t ). Solving this gives t = 2 or t = 1 2 . The slope of the focal chord PQ is m = at^2 - a t^2 2at - (- 2a t ) = t - 1 t 2 . Since PQ makes an acute angle with the