Question

Length of shortest normal chord of parabola

Answer 1

Answer:

Step-by-step explanation:

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  

since here a = 1