Q.53. Minimum length of normal chord of a parabola

Question 53.
Find the least length of the normal chord intercepted by the parabola y^2 = 4ax.

Answer 53.
Let the normal chord at P(at^2, 2at) to the parabola be of length r
If the slope of the normal is tanθ, the other endpoint of the chord is
Q (at^2 + rcosθ, 2at + rsinθ)
As it lies on the parabola,
(2at + rsinθ)^2 = 4a(at^2 + rcosθ)
=> 4a^2t^2 + 4atrsinθ + r^2 sin^2 θ = 4a^2t^2 + 4arcosθ
=> rsin^2 θ = 4a (cosθ - tsinθ) ... (1)

y^2 = 4ax
=> dy/dx = 2a/y
and slope of normal, tanθ = - dx/dy = - y/2a = - t

Putting t = - tanθ in (1),
=> rsin^2 θ = 4a (cosθ + sin^2 θ/cosθ)
=> rsin^2 θ cosθ = 4a
=> chord length, r = 4a/(sin^2 θ cosθ)
=> r = 4a / [(1 - u^2) u] (putting cosθ = u)

For r to be minimum, (1 -u^2) u should be maximum
=> 1 - 3u^2 = 0 => u = 1/√3
Also, second derivative of (1 - u^2) u is - 6u < 0
=> (1 - u^2) u is maximum at u = 1/√3.
=> Minimum chord length,
r = 4a / [(2/3)(1/√3)]
= 6√3 a

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