Find a length of base and leg of minimum area isosceles triangle circumscribed about a ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ 
Find a length of base and leg of minimum area isosceles triangle circumscribed about a ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ if its base is parallel to x-axis

Similar problem, where I am given circle instead of ellipse is easy, because I can use similar triangles to write leg in terms of base and diameter of circle (or vice versa), but in this case I have no idea because I don't know what's the distance from center of ellipse to leg of triangle.
I thought about writing an eqation of ellipse in parametric form but that didn't seem to help me much.
How should I solve this problem?
 A: In the figure below, let $F=(0,h)$ and $E=(x_0,y_0)$, with $FE$ tangent to the ellipse.

Then $$\frac{x_0^2}4+\frac{y_0^2}9=1$$
The slope of the tangent is given by the derivative:
$$y'=\left(3\sqrt{1-\frac{x^2}4}\right)'=-\frac34\frac x{\sqrt{1-\frac{x^2}4}}=-\frac94\frac xy$$
You could have gotten this result with using implicit derivative.
Then, at $E$ the slope of the tangent is equal to the slope of the $FE$ line.
$$\frac{y_0-h}{x_0}=-\frac94\frac{x_0}{y_0}$$
Rearranging the terms:
$$\frac{y_0^2}9-\frac{y_0h}9+\frac{x_0^2}4=0$$
or $$y_0=\frac9h$$
From here, calculate $x_0$ in terms of $h$, then the $x$ coordinate of $G$ (knowing that the $y$ coordinate is $-3$). Then the area of the triangle is $x_G(h)\times (h+3)$. Take the derivative, set it to zero, and you get $h$.
A: 
The circle equal area with ellipse has radius R and we have:
$A=2\times 3 \pi=\pi R^2\Rightarrow R=\sqrt 6$
We know the triangle inscribed in circle has maximum area when it is equilateral, so we have:
$BC=a=R\cos 30^o\times 2=\sqrt 6\times \frac {\sqrt 3}2=\frac 3{\sqrt 2}$
$AH=h=a\sin 60^o=R\sqrt 3\times \frac {\sqrt 3}2 R=\frac {3\sqrt 6}2$
$AH'=h'=\frac 2{\sqrt 6}\times h=3$
$x_{H'}=OH'=AH'-AO=3-\sqrt6=0.55$
Putting this in equation of ellipse we get:
$B'H'\approx 2.884$
Finally the maximum area is :
$$A_{AB'C'}=h' \times B'H'\approx 8.64$$
A: Like the figure in @Andre, the eq. of the base is $y=-3$, the eq. of tangent is
$y=-mx + d, m>0, d=\sqrt{4m^2+9}.$ Three vertices of triangle are $(0.-3), (0,d),((d+3)/m,-3)$. The area of the full triangle is
$$A=\frac{(d+3)^2}{m} \implies \frac{dA}{dm}=\frac{(d+3)}{m^2}(8m^2-d^2-3d)$$
$\frac{dA}{dm}=0$ can be re-written using $z=4m^2$ as $$(z-9)^2=9(z+9)=0 \implies z=27 \implies m=3\sqrt{3}/2$$
Finally, we get the height of triangle as $h=d+3=\sqrt{z+9}+3=9$ and the breath of full triangle is $b=2(d+3)/m=4\sqrt{3}$ corresponding to $A_{min}=18\sqrt{3}.$
