Maximising the area of a triangle with its vertices on a parabola.

Question

Let $A\equiv (4,-4)$ and $B\equiv(9,6)$ be tw points on the parabola $y^2 = 4x$. Let $C$ be a point on the parabola between $A$ and $B$ such that the area of the triangle $\triangle ABC$ is maximal. What are the coordinates of $C$?

The parametric coordinates of C are $(t^2 , 2t)$. Using Shoelace formula, I get the area of the triangle to be $5t^2 -5t$. But this has no global maximum? In this case, what shluld I do?

Answer 1

The shoelace formula for the points $(4, -4), (9,6), (t^2, 2t)$ gives $$ A = \frac12\left|4\cdot 6 + 9\cdot 2t + t^2\cdot (-4) - (-4)\cdot 9 - 6\cdot t^2 - 2t\cdot 4 \right|\\ = \frac12|24 + 18t - 4t^2 + 36 - 6t^2 - 8t|\\ = |30 + 5t - 5t^2| $$ Since we "[l]et $C$ be a moving point on the parabola between $A$ and $B$", that means that $t$ is between $-2$ (which makes $C = A$) and $3$ (which makes $C = B$). On this interval, $30 + 5t - 5t^2$ is positive, so we can drop the absolute value signs, and find a maximum at $t = 0.5$, which gives $C = (\frac14, 1)$ and area of $31.25$.

Answer 2

The equation of $AB$ is $y=2x-12$.

Let $l||AB$.

We see that $S_{\Delta ABC}$ would be maximal for $C\in l$ and $l$ is a tangent line to the parabola.

Let $C\left(\frac{c^2}{4},c\right)$.

Thus, $yc=2\left(x+\frac{c^2}{4}\right)$ is an equation of the tangent line,

which gives $\frac{2}{c}=2$, $c=1$ and $C\left(\frac{1}{4},1\right)$.