Question

Prove that the point on the parabola y² = 4ax (a > 0) nearest to the focus is vertex. 

Answer 1

. Let P(at² ,2at) be the point on the parabola. 

y² = 4ax, which is nearest to the focus S(a, 0) then 

sp² = (at² – a)² + (2at – 0)² 

 f(t) = a² 2(t² – 1)(2t) + 4a² (2t) 

 = 4a² t(t² – 1 + 2) = 4a² t(t² + 1) 

For minimum value of f(t) = 0 ⇒ t = 0 

 f″(L) = 4a² (3t² + 1) 

 f′(0) = 4a² > 0 

∴ At t = 0, f(t) is minimum 

Then P = (0, 0) 

∴ The point on the parabola y² = 4ax, which is nearest to the focus is its vertex A(0, 0)