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Find the area of the triangle formed by the tangents from the point $(h, k)$ to the parabola $y^2=4ax$ and the chord of contact.

I find the the tangents from external points $(h,k)$, i.e. $$(y^2-4ax)(k^2-4ah)=(ky-2a(x+h))^2$$ Again the chord of contact of the given parabola is $ky=2a(x+h)$. Then we try to find out three extremities.

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  • $\begingroup$ What are your thoughts? Please show your work first. $\endgroup$
    – Deepak
    May 25 '17 at 3:03
  • $\begingroup$ I find the the tangents from external points $(h,K)$, i.e. $(y^2-4ax)(k^2-4ah)=(ky-2a(x+h))$. again the chord of contact of the given parabola is $ky=2a(x+h)$. Then we try to find out three extremites $\endgroup$
    – MSMM
    May 25 '17 at 3:11
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Let $(x_1,y_1)$ and $(x_2,y_2)$ be the points of contact.

Substitute the chord $ky=2a(x+h)$ into the parabola $y^2=4ax$ to obtain a quadratic equation in $y$. $y_1$ and $y_2$ are the roots. So we have $y_1+y_2$ and $y_1y_2$. $(y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2$.

$k(y_1-y_2)=2a(x_1-x_2)$ and so we can find the distance between the two points of contact.

The distance between the point $(h,k)$ and the chord of contact is $\displaystyle \left|\frac{2a(h)-k(k)+2ah}{\sqrt{4a^2+k^2}}\right|$.

Multiply the two distances and divide it by $2$, we have the area.

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