Question

Y2=8x ka focus s and pq is a common chord of it and x2+y2-2x-4y=0 .Then area of triangle pqs

Answer 1

We know that the normal of a parabola is y = mx – 2am – am3.

Hence, the equation of normal for y2 = x is y = mx – m/2 – m3/4

Since the normal passes through (c, 0), we have

mc – m/2 – m3/4 = 0

This gives m(c – 1/2 – m2/4) = 0

This yields either m = 0 or m2 = 4(c-1/2)

For m = 0, the equation of normal is y = 0.

Also, m2 ≥ 0, so c-1/2 ≥ 0 or c ≥ ½.

At c = ½, we have m = 0.

For the other normals to be perpendicular to each other, we must have m1m2 = -1.

m2/4(1/2 – c) = 0 has m1m2 = -1.