Maximise area of ellipse inscribed inside an isosceles triangle 
Find the maximum area(in terms of $A$) of the ellipse inscribed inside an isosceles triangle of area $A$, having one axis lying along the perpendicular from the vertex of the triangle to its base.

WLOG, Let the vertices of the isosceles triangle be $A(p,0),B(0,q),C(0,-q)$
$A=1/2*p*2q=pq$.
Let the centre of the ellipse be $(a,0)$, Let the ellipse have axis of lengths $2a,2b$.
The equation of the inscribed ellipse is $\frac{(x-a)^2}{a^2}+\frac{y^2}{b^2} = 1$.
The ellipse has tangent equations $\frac{x}{p} \pm \frac{y}{q}=1$.
General tangent equation for ellipse(with shifted centre, centre at(a,0)) is
$y=m(x-a) \pm \sqrt{a^2m^2+b^2}$.
So relating any of the lines with general equation, 
Gives $q(p-a)^2=a^2q^2+b^2p^2$,
Or $pq^2-2aq=bp^2$.
Or $A^2-2aA=bp^3$
So area of ellipse=$\pi ab=
\pi a \frac{A^2-2aA}{p^3}$.
However I'm unable to eliminate p in this. 
How to proceed?
 A: SteinerInellipse:

"The Steiner inellipse of a triangle has the largest area of any
  inellipse of that triangle; its area is $\tfrac\pi{3\sqrt3}$ times the
  area of the triangle".

The Steiner inellipse
is tangent to the sides of the triangle at their midpoints,
so for the isosceles triangle one of its axes 
will be localed along the perpendicular from the vertex of the triangle to its base,
hence it is indeed a solution.

For the isosceles $\triangle ABC$ with side lengths 
$|AC|=|BC|=a$ and the base $|AB|=c$
the lengths of semi-axes 
of the Steiner inellipse are
\begin{align} 
s_a&=
\max\Big(\tfrac{\sqrt3}6\,c,\, \tfrac16\,\sqrt{4a^2-c^2}\Big)
,\\
s_b&=
\min\Big(\tfrac{\sqrt3}6\,c,\, \tfrac16\,\sqrt{4a^2-c^2}\Big)
.
\end{align} 

Update:

Let WLOG $A=(-1,0)$, $B=(1.0)$, 
$C=(0,q)$, 
$s_a$ and $s_b$-
the major and minor semi-axes of the ellipse, respectively.
Then 
the center of the ellipse $O=(0,s_a)$
and
the equation of the ellipse is
\begin{align} 
\frac{x^2}{b^2}+\frac{(y-a)}{a^2}&=1
\tag{1}\label{1}
,
\end{align}
and the equation for the upper arc is
\begin{align}
y(x)&=\frac{s_a}{s_b}\cdot\Big(s_b+\sqrt{s_b^2-x^2}\Big)
\tag{2}\label{2}
,\\
y'(x)&=
-\frac{s_a}{s_b}\cdot \frac{x}{\sqrt{s_b^2-x^2}}
\tag{3}\label{3}
.
\end{align}
Equation for the tangent line $AC$ is
\begin{align}
y_t(x)&=q\,x+q
\tag{4}\label{4}
.
\end{align}
Let the tangent point be $B_m=(x_0,y_0)$.
Then 
\begin{align}
y'(x_0)&=q
,\\
x_0&=
-\frac{q\,s_b^2}{\sqrt{q^2\,s_b^2+s_a^2}}
.
\end{align}
Also
\begin{align}
y_0 &= q\,x_0+q
,\\
y_0&=
\frac{s_a}{s_b}\cdot\Big(s_b+\sqrt{s_b^2-x_0^2}\Big)
.
\end{align}
This is enough to express $s_a,\,s_b$ in terms of $x_0$:
\begin{align}
s_a&=q\cdot\frac{1+x_0}{2+x_0}
,\\
s_b&=\sqrt{-\frac{x_0}{2+x_0}}
.
\end{align}
The area of the inellipse is then
\begin{align}
S_e(x_0)&=\pi\cdot s_a\cdot s_b
\\
&=
q\cdot \pi\,(1+x_0)\cdot\sqrt{-\frac{x_0}{(2+x_0)^3}}
,\\
S_e'(x_0)&=
-\frac{q\cdot\pi\cdot(1+2\,x_0)}{\sqrt{-x_0\,(2+x_0)^5}}
,
\end{align}
and the maximal area is reached at $x_0=-\tfrac12$,
as expected.
