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.

							not possible
						

							
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’.
 
On solving the equation of these two points we will arrive at, (hint: factorise denominator of slope)
 
(t – 4/t)y – 2at2 – 8a = 2x – 2at2
(2x – 8a) – (t – 4/t)y = 0.
 
This line clearly goes through the point (-4a, 0) which lies on the axis of the parabola y = 0.