A classical minimization problem

If we translate the trapezoid over $(-5,0)$ , the y-axis bexomes the vertical axis of symmetry. Let $y=y_0$ be the horizontal axis of symmetry, then we have the following equation for the ellipse: $\frac{x^2}{a^2} + \frac{(y-y0)^2}{b^2}=1$ Filling in the coordinates of the two known points $(x,y)=(5,0)$ and $(x,y)=(2,5)$ gives two equations that must hold: $\frac{25}{a^2}+\frac{y_0^2}{b^2}=1 \tag{I}$ $\frac{4}{a^2}+\frac{(5-y_0)^2}{b^2}=1 \tag{II}$

Subtract $4b^2$ times (I) from $25b^2$ times (II) and rearrange to get $b^2=y_0^2-\frac{250}{21}y_0+\frac{625}{21}$

and rewrite (I) as $a^2=\frac{25b^2}{b^2-y_0^2}=25\frac{21y_0^2-250y_0+625}{625-250y_0}$

Now since the quantity $a^2b^2$ is a constant times the area squared ( $A=πab$ ), at minimum area it must be stationary. $a^2b^2=\frac{(21y_0^2-250y_0+625)^2}{21(25-10y_0)}$ Using the quotient and chain rules we set the derivative to 0:

$\frac{d 21 a^2b^2}{dy_0} =\frac{2(...)(42y_0-250)(25-10y_0)+10(...)^2}{(25-10y_0)^2}=0$

where (...) is shorthand for $21y_0^2-250y_0+625$ . Because we demand that $b \neq 0$ we can divide by (...) and simplify to find

$63y_0^2-460y_0+625=0$ This gives two values for $y_0$ . The larger value would give $b^2<0$ , so we take the smaller root $y_0=\frac{460-\sqrt{460^2-4×63×625}}{126}=1.80480899...$ From $y_0$ we calculate $A=π\sqrt{a^2b^2}=62.974613...$ and find the answer $\lfloor{100A\rfloor=\boxed{6297}}$