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Tangent of coordinate (a.t^2, 2at)
x= a.t^2 y = 2at
dx/dt = 2at dy/dt = 2a
dy/dx = dy/dt * dt/dx = 2a/2at = 1/t.
The Equation of Tangent of the Parabola is
y – 2at = (1/t)(x – a.t^2)
à ty – 2.a.t^2 = x – a.t^2
à ty = x + a.t^2
The Point of Intersection of Two Tangents of a Parabola
à t1.y – 2.a..t1^2 = x – a.t1^2
à t2.y – 2.a..t2^2 = x – a.t2^2
Subtracting
à y(t1 – t2) = a(t1^2 – t2^2)
à y = a(t1 + t2)
it follows that x = at1(at1.t2 – a.t^2) = a.t1.t2
I hope, Now you can do it
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