Given:
The parabola, y2 = 4px and a double ordinate of length 8p.
Let AB be the double ordinate of length 8p for the parabola y2 = 4px.
Now, let us compare to the actual parabola, y2 = 4ax
Then,
axis is y = 0
vertex is O(0, 0).
We know that double ordinate is perpendicular to the axis.
So, let us assume that the point at which the double ordinate meets the axis is (x1, 0).
Then the equation of the double ordinate is y = x1. It meets the parabola at the points (x1, 4p) and (x1, – 4p) as its length is 8p.
Now, let us find the value of x1 by substituting in the parabola.
(4p)2 = 4p(x1)
x1 = 4p.
The extremities of the double ordinate are A(4p, 4p) and B(4p, – 4p).
Assume the slopes of OA and OB be m1 and m2. Let us find their values.
m1 = (4p – 0)/(4p – 0)
= 4p/4p
= 1
m2 = (4p – 0)/(-4p – 0)
= 4p/-4p
= -1
So, m1.m2 = 1. – 1
= – 1
The product of slopes is -1. So, the lines OA and OB are perpendicular to each other.
Hence the extremities of double ordinate make right angle with the vertex.