Question

Prove that the locus of the midpoints of chords of a parabola which subtend right angle at the vertex is another parabola whose latus rectum is half that of the original parabola. Further, all such chords pass through a fixed point on the axis of the parabola.

Answer 1

See Fig. Let M(x₁, y₁) be the midpoint of a chord of S ≡ y² - 4ax = 0 which subtends a right angle at the vertex. Hence , equation of the chord is

Suppose the chord provided in Eq. (1) cuts the parabola at points P and Q. Therefore, the combined equation of the pair of lines (bar)OP and (bar)OQ is

Since ∠POQ = 90º , in Eq. (2)

oefficient of x² + Coefficient of y² = 0
 

Hence, the locus of (x₁, y₁) is the parabola y² = 2a(x - 4a). If we put y₁² = 2ax₁ - 8a² in Eq. (1), we get the equation of the chord as yy₁ - 2ax = -8a² which passes through (4a, 0).