# Maximum area of a triangle inside an ellipse.

The question says to find the maximum area of a triangle formed by joining the points $A,B$ and origin $O$, where $A$ and $B$ are points of intersection of an arbitrary line passing through $(4,5)$ with the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$.

I tried to parametrize the line and then solve it with the curve, find the points of intersection and maximize the area, but this method is not suitable as it's too lengthy. Is there a better way to approach the problem?

-

If the aim is just to find the area of the largest triangle, rather than any details, I would approach it this way:

Consider a circle of radius $r$ and so area $\pi r^2$. The largest triangle in the circle with one vertex at the centre is the right-angled triangle with the right angle at the centre and with area $r^2 /2$. So the ratio of the area of the circle to the area of the largest triangle with one vertex at the centre is $2\pi$.

The same $2\pi$ ratio applies to the area of an ellipse compared to the area of the largest triangle with one vertex at the centre, since this is just a linear transformation of the circle.

The area of your ellipse is $6\pi$ so the area of the largest triangle will be $3$. There will in fact be two such triangles, one each side of the line joining $(4,5)$ to the origin.

-
As a warning, the largest triangles in the ellipse need not be right angled – Henry Jul 19 '14 at 17:28
Hence, somewhat surprisingly, the answer is independent of the particular external point chosen ($(4,5)$ in this case). – TonyK Jul 19 '14 at 22:20
@TonyK: It is true and nice: the maximum area is always $\frac{1}{2\pi}$ times the area of the ellipse, hence $\frac{ab}{2}$. – Jack D'Aurizio Jul 20 '14 at 0:24

I read this question too late, anyway I completely agree with @Henry's approach: scale, solve the circular case, scale back. As a proof without words:

-