You may use the fact that the product of perpendicular distances of any tangent on ellipse from its foci is
always equal to square of semi-minor axis.
Since the path of centre is x2
= 5 when y = 1, x = 2.
The incident ray passes through origin and (2, 1) ⇒ its equation is y = 1/2x
Given parabola is (y – 5)2
= – 4 (x – 10)
whose vertex (10, 5) lies on y = 1/2x.
The tangent at the vertex is the line x = 10.
Figure shows that OP = 2 (OR) = 10
so reflected ray cuts y-axis at (0, 10)