Answer to Question #124564 in Geometry for Tony

Question

f a chord of the parabola y^2 = 4ax is a normal at one of its ends, show that its mid-point lies on the curve 2(x − 2a) = y^2/a + 8a^3/y^2

Answer 1

Equation of the normal chord at any point (at₂,2at) of the parabola is

y+tx=2at+at³ .....(1)

Equation of the chord with midponit (x₁,x2₂) is T=S1

or yy₁-2a(x+x₁)=y₁²-4ax₁ or

yy₁-2ax=y₁²-2ax₁ ....(2)

since equation (1)and(2) are identical

1/y₁ = t/(-2a) = (2at+at³)/t = 2a + {(-2a)/y₁}²

or -(y₁²)/2a+x₁ = 2a+ 4a³/(y₁²)

or x₁-2a = (y₁²)/2a +4a³/(y₁²)

hence the locus of the middle point (x₁,y₁) is

x-2a = y²/2a + 4a³/y²

multiply 2 on both sides

2(x-2a) = 2y²/2a + 2(4a³/y²)

2(x-2a) = y²/a + 8a³/y²

hence midpoint lies on this curve