geometrytangent line
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.
You can easily show that tangents to $y^2=4ax$ at points $t_1$ and $t_2$ intersect at the point $(at_1 t_2, a(t_1 + t_2))$. Now we are given three points $t_1, t_2, t_3$ satisfying the two relations:
$$(at_1 t_2)^2=4ab(t_1 + t_2)$$ $$(at_3 t_2)^2=4ab(t_3 + t_2)$$
And we need to find the locus of the intersection of other two tangents, that is, a relation between $x=at_1 t_3$ and $y=a(t_1 + t_3)$.
Note that, by rearranging the above equations, both $t_1$ and $t_3$ satisfy the quadratic equation $t^2(at_2^2) - 4bt - 4bt_2=0$ (assuming $t_2$ to be constant). Thus we can find $t_1 t_3$ and $t_1 + t_3$ directly from this equation which are the sum and product of roots of the quadratic. Thus:
$$t_1 + t_3 = \frac{4b}{at_2^2}$$ $$t_1t_3 = \frac{4b}{at_2}$$
Now you can substitute $x=at_1 t_3$ and $y=a(t_1 + t_3)$ in the above relations, and straightforward manipulation leaves you with the locus $x^2=4by$.
HINT.
From any point $A$ of the hyperbola, in the branch intersecting the parabola but external to it (see diagram), draw tangents $AD$ and $AF$ to the parabola, which intersect the other branch of the hyperbola at $B$ and $C$. Check that line $BC$ is also tangent to the parabola.
Best Answer
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.