Show that all chords of a parabola which subtend a right angle at the vertex pass through a fixed point on the axis of the curve.

Let’s say that the parabola is – y² = 4ax.

 

Let the ends of the chords on the parabola be A(at₁², 2at₁) and B(at₂², 2at₂). Now as they subtend a right angle at the vertex of the parabola - m₁m₂ = -1, where m₁ is slope of line joining AO and m₂ is slope of line joining BO, where O = (0,0) or origin.

 

Then, as follows – (at₁² – 0) * (at₂² – 0) + (2at₁ – 0) * (2at₂ – 0) = 0, as x₁x₂ + y₁y₂ = -1. When we solve this equation we will get the relation between t₁ and t₂ as follows -

 

t₂ = -4 / t₁ and let t₁ = t, so then the ends of the chords become A(at², 2at) and B(16a / t², -8a / t). We can get the equation of the line passing through the ends of chords or just the equation of the chord ‘AB’.

 

 

 

This line clearly goes through the point (-4a, 0) which lies on the axis of the parabola y = 0.