Conics and calculus

Let $ABC$ be an isosceles triangle such that $AB=AC$ . Let $AO \bot BC$ be the altitude. Then according to the question $BC=AO=2$ . Let $OLMK$ be the required ellipse having center at ${ O }^ { ' }$ such that $PQ=2a$ and $MO=2b$ . Extend $OA$ to $Y$ and $OB$ to $X$ respectively. Let ${ X }^{ ' }OX$ and ${ Y }^{ ' }OY$ be the x-axis and y-axis respectively. Then the equation of ellipse will be: $\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { \left( y-b \right) ^{ 2 } }{ { b }^{ 2 } } =1$ and equation of $AB$ will be: $\frac { x }{ 1 }+ \frac { y }{ 2 }=1 \Rightarrow y=-2x-2 \quad\quad {\because OB=2}$ Now since $AB$ is tangent to ellipse at $L$ . It's equation must be of the form $y-b=mx\pm \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } }$ Comparing we get , $m=2$ and $b\pm \sqrt{ {4 a }^{ 2 }+{ b }^{ 2 }=2 } \Rightarrow b=1-{ a }^{ 2 }$

Now area of ellipse= $\pi ab$ {standard result}

$A=\pi a(1-{ a }^{ 2 })$

Now At $A$ = ${A}_{max}$ $\frac {dA}{da}=0$

or $\pi (1-3{ a }^{ 2 })=0 \Rightarrow a=\pm \frac{ 1 }{ \sqrt{ 3 } }$

Substituting this in previous equation , we get ${ A }_{ max }= { \frac{ 2\pi }{ 3\sqrt{ 3 } } }$ Hence, $A=2$ , $B=C=3$ . So $A+B+C=2+3+3=\boxed{ 8 }$