Optimize triangle on ellipse

Question

So I have two points A(3,1) and B(5,0) on the ellipse $x^2+16y^2=25$. I am asked to find the third point C on the same ellipse that maximizes the area of the triangle ABC. So I know I have to find the maximum of $x^2+16y^2=25$ with a constraint. That is where I am confused. I think that the constraint should be the optimized version of the area equation of the triangle. So here is the work I have...

|$\overline{AB}$| = $\sqrt{(5-(-3))^2+(0-1)^2}$ = $\sqrt{65}$

A = $\frac12*b*h$

A = $\frac12*\sqrt{65}*h$

Now how do I make this a constraint, or am I not going about this right?

Answer 1

HINT:

WLOG the third point be $C\left(5\cos t,\dfrac54\sin t\right)$

The equation of $AB$ $$\dfrac{y-0}{x-5}=\dfrac{0-1}{5-3}\iff x+2y-5=0$$

The length of $A(3,1);B(5,0)$ is constant, we need to maximize

the perpendicular distance of $C$ from $AB$

which is $$\dfrac{|1\cdot5\cos t+2\cdot\dfrac54\sin t-5|}{\sqrt5}= \dfrac1{\sqrt2}|2\cos t+\sin t-2|$$

Now $2\cos t+\sin t=\sqrt5\cos\left(t-\arccos\dfrac2{\sqrt5}\right)$

$-1\le\cos\left(t-\arccos\dfrac2{\sqrt5}\right)\le1$

$\iff-\sqrt5-2\le2\cos t+\sin t-2\le\sqrt5-2$

$-\sqrt5-2\le2\cos t+\sin t-2,\implies(\cos t+\sin t-2)^2\le(-\sqrt5-2)^2=(\sqrt5+2)^2$

Answer 2

One way to do this is to observe that under a suitable scaling transformation of the coordinate plane, areas remain proportional; thus if we scaled one coordinate axis in a way that makes the ellipse a circle, then in the scaled coordinate system, the triangle with the maximal area is isosceles.

Since the semimajor axis is $5$ and the semiminor axis is $5/4$, one convenient transformation is to let $y' = 4y$. This results in the ellipse becoming the circle $$x^2 + (y')^2 = 5^2.$$ Then $A' = (3,4)$ and $B' = (5,0)$. Hence $C'$ must satisfy $A'C' = B'C'$. To locate this point, we find the midpoint $M = (4,2)$ of $A'B'$, thus $C'$ is collinear with $(0,0)$ and $M$. Two such candidates exist on the circle: $$C' \in \left\{ \pm \left(\frac{4(5)}{\sqrt{4^2+2^2}}, \frac{2(5)}{\sqrt{4^2+2^2}} \right) \right\} = \left\{(2\sqrt{5}, \sqrt{5}), (-2\sqrt{5},-\sqrt{5})\right\}.$$ Since $A'$ and $B'$ are in the same (first) quadrant, we choose $C'$ in the third quadrant. Transforming back, we obtain $C = (-2\sqrt{5}, -\sqrt{5}/4)$. From here, it is also trivial to find the maximum area; we simply calculate it in the transformed system and then multiply it by $4$.