Area of triangle inscribed in a parabola How can u prove that the area of the triangle inscribed in a parabola is twice the area of the triangle formed by the tangents at the vertices?
 A: Claim: Given any 2 points on the parabola, the area that is bounded by the line between these 2 points and the parabola, is twice the area that is bounded by the 2 tangents and the parabola.
Proof. Standard calculus techniques work.
WLOG, we normalize the quadratic to be of the form $ y = x^2$. Let the points be $ (a, a^2), (b, b^2)$ with $a<b$. The equation of the tangents are $ y = 2ax - a^2$ and $ y = 2bx -b^2$. They intersect at $x = \frac{ b+a} { 2}, y = ab$.
The area between the two points and the parabola is equal to the trapezium minus the area under the curve. The area of the trapezium is $\frac{ a^2 + b^2}{2} \times (b-a)$. The area under the curve is $\int_a^b x^2 \, dx = [\frac{1}{3} x^3]_a^b = \frac{ b^3 - a^3 } { 3}$. Hence, the area is $ \frac{b^3-ab^2+a^2b-a^3}{2} - \frac{b^3-a^2}{3} = \frac{ b^3-3ab^2+3a^2b-a^3 } { 6}$.
The area between the two points and the intersection of their tangents are given by 
$ 2ab^2 + \frac{b^3}{2} + \frac{ab^2}{2} -2ab^2 - \frac{ a^3}{2} - \frac{ a^2b}{2} = \frac{ b^3-3ab^2+3a^2b-a^3 } { 2}$. This is thrice of the area that we calculated above. Hence, the area that is between the parabola and the tangents, is twice the area between the parabola and the line.
Corollary: Apply the above claim to each pair of your 3 points. We get that the area of the triangle within the parabola is twice the area of the triangle formed by tangents.
A: Let the $3$ points be in parametric form : $t_1,t_2,t_3$
where $t\equiv (at^2,2at)$
I am assuming parabola $y^2=4ax$
You can find this area by determinant formula. It will come out to be :
$$a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)$$
The tangents at $t_i,t_j$ intersect at $at_it_j,a(t_i+t_j)$
Why : Equation of tangent is $yy_1=4a(x+x_1)/2$ at $(x_1,y_1)$ You can put parametric $(x_1,y_1)$ and find point of intersection of 2 tangent lines.
Use determinant formula again to compute area:
$$\frac 1 2a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)$$
