Q.56. Parabola, gradient of a chord whose midpoint lies on a given line

Question 56.
Show that the midpoint of a chord of the parabola x = 2at, y = at^2 lies on the vertical line x = k if and only if the chord has gradient k/2a.

Answer 56.
Eliminating t, equation of the parabola can be written as x^2 = 4ay.
Since the midpoint lies on the line x = k, its x-coordinate is k.
Let the y-cordinate be m (m is variable, meaning having different values for different chords).
Thus, let midpoint be (k, m).

Let the endpoints of the chord be at a distance r from the midpoint.
If the gradient of the chord is tan Φ, then the end points are
(k + rcos Φ, m + rsin Φ) and (k - rcos Φ, m - rsin Φ)

 As these endpoints lie on the parabola,
(k + r cos Φ)^2 = 4a (m + r sin Φ) ... ( 1 ) and
(k - r cos Φ)^2 = 4a (m - r sin Φ) ... ( 2 )

Subtracting equation ( 2 ) from ( 1 ),
4krcos Φ = 8arsin Φ
=> tan Φ = k / 2a
=> gradient = k / 2a.

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