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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
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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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