A parabola has focus $F$ and vertex $V$, where $VF = 10$. Let $AB$ be a chord of length $100$ that passes through $F$. Determine the area of triangle $V\!AB$.
This is an olympiad question which I came across last week. I really don't have any idea where to start. I think the information provided in the question is not even enough to solve the problem.
I only know that for a parabola $y^2 = 4ax$, the length of the focal chord through $t$ is given by $a\left(t+\dfrac1t\right)^2$.
Can anyone check the problem if it's correct? If yes, then how may I proceed? Any hint would be enough.
Best Answer
https://en.wikipedia.org/wiki/Parabola
The polar equation of parabola is $$r=\frac p{1-\cos\varphi} \tag 1 \label 1$$ where
The focal length in turn is the distance from the focus of a parabola to its vertex, and it is given as $f = VF = 10.$
The endpoints of a chord, which is rotated by $\varphi$ from the parabola's axis, are at the distances given by the parabola equation $\eqref 1$: $$\begin{cases} r_1 = FA = \frac p{1-\cos\varphi} \\ r_2 = FB = \frac p{1-\cos(\varphi+\pi)} = \frac p{1+\cos\varphi} \end{cases}$$ Now, the length of the chord AB, a base of our triangle $\triangle ABV$, is: $$AB = r_1+r_2 = \frac p{1-\cos\varphi} + \frac p{1+\cos\varphi} \\ = p\,\frac{(1+\cos\varphi)+(1-\cos\varphi)}{(1-\cos\varphi)(1+\cos\varphi)} \\ = \frac{2p}{1-\cos^2\varphi} = \frac{4f}{\sin^2\varphi} \tag 2 \label 2$$ On the other hand, the height of the triangle, i.e. the distance of the vertex V from the line AB, is: $$h = VF\,\sin\varphi = f\sin\varphi$$ From $\eqref 2$ we get: $$\sin\varphi = \sqrt{\sin^2\varphi} = \sqrt{\frac{4f}{AB}}$$ so the area of the triangle $$\frac 12 h\cdot AB = \frac 12 f\sin\varphi\cdot AB = \frac 12 f\cdot AB\cdot\sqrt{\frac{4f}{AB}} $$ $$\boxed{ P_{\triangle ABV} = f\cdot\sqrt{f\cdot AB}}$$ Given $f=10$ and $AB=100$ we get: $$ P_{\triangle ABV} = 100\sqrt{10}.$$