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I have no clue how to work it out or to draw it. If you can help please do. 

 Apr 24, 2019

The parabola has its vertex at the origin and opens up to the right with the positive x=axis as its axis of symmetry.

Let the point Q have the standard parametric co-ordinates (at^2, 2at).

The chord passes through the focus (a, 0), so work out its equation, y(t^2 - 1) = 2t(x-a), and then where it cuts the parabola again,

at x = a/t^2, y = -2a/t, (y negative assuming that we chose Q to lie above the x-axis). 

Calculate the equation of the tangent at Q, (its slope will be 1/t) and the equation will be ty = x + at^2.

The line PT will have the equation y = -2a/ t, 

so eliminate the parameter t between these last two equations, and the result (x + 2a)y^2 + 4a^3 = 0 drops out.

 Apr 25, 2019

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