Question

Find the locus of the middle points of chords of a parabola which subtend a right angle at the vertex, and prove that these chords all pass through a fixed point on the axis of the curve.

Answer 1

Let PQ be any such chord, and let its equation be

y = mx + c .....(1)

The lines joining the vertex with the points of intersection of this straight line with the parabola

y2 = 4ax .........(2)

are given by the equation

y2c = 4ax(y - mx).

These straight lines are at right angles if

c + 4am = 0.

Substituting this value of c in (1), the equation to PQ is

y = m(x - 4a) .......(3)

This straight line cuts the axis of a; at a constant distance 4a from the vertex, i.e. AA' = 4a.

If the middle point oi PQ be (h, k) we have,

k = 2a/m ..............(4)

Also the point (h, k) **** on (3), so that we have

k = m(h - 4a) ..........(5)

If between (4) and (5) we eliminate m, we have

so that (h, k) always **** on the parabola

y2 = 2a(x - 4a).

This is a parabola one half the size of the original, and whose vertex is at the point A' through which all the chords pass.