The chord of contact of two tangents drawn from an external point P to the parabola x^2=8y has equation x+2y-3=0. Find the coordinates of P

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given the parabola `x^2=8y` and the chord of contact `x+2y-3=0` of two tangents drawn from an external point P, find the coordinates of P.

Using parametric equations:

(1)The equation of the chord of contact is `y-1/2x(p+q)+apq=0` *

Rewrite `x+2y-3=0` in this form:

`2y+x-3=0 ==>y+1/2x-3/2=0`

So `-1/2x(p+q)=1/2x==>p+q=-1` and `apq=-3/2` by equating corresponding terms.

(2) The equation of the parabola in parametric form is `y=x^2/(4a)` .

So `x^2=8y ==>y=x^2/8==>y=x^2/((4)(2)) ==>a=2`

(3) The point of intersection of the tangents is given by

`(a(p+q),apq)` **

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We have `a=2,p+q=-1,"and"apq=-3/2` so the point of intersection is `(2(-1),-3/2)` or `P=(-2,-3/2)`

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The graphs:

Sources:

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