A very hard problem indeed

The general equation for an ellipse is $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ where $B^2-4AC<0$ .

Since this ellipse passes through $(0,0)$ , we must have $F=0$ . Without loss of generality, we can take $A=1$ , to leave the equation as $x^2+Bxy+Cy^2+Dx+Ey=0$

The point $(10,0)$ tells us that $100+10D=0$ ie that $D=-10$ .

Updating our equation, $x^2+Bxy+Cy^2-10x+Ey=0$

The other two points tell us that $\begin{aligned} 49+42B+36C-70+6E&=0 \\ 9+21B+49C-30+7E&=0 \end{aligned}$

We have two equations in three unknowns; although this means we can't solve the system completely, we can parameterise two of the unknowns in terms of the third. Choosing (arbitrarily) $B$ as the parameter, we get $C=4B-\frac12,\;\;E=\frac{13}{2}-31B$

So (almost) the whole family of conics passing through the given points are parameterised as $x^2+Bxy+\left(4B-\frac12\right)y^2-10x+\left(\frac{13}{2}-31B\right)y=0$

This form in fact includes conics that are not ellipses, but we'll deal with those later. It also excludes conics where (in the original general form) $A=0$ ; however, these cannot be ellipses, so - for this problem - we're not interested in them.

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Now we turn to the area enclosed. We can work out expressions for the semi-major and semi-minor axes of the ellipse ( $a$ and $b$ ) in terms of the coefficients $A$ to $F$ ; or we can look them up, eg here .

The area of the ellipse is $\pi ab$ . This works out as $T=\pi ab=\frac{2\pi \left(AE^2+CD^2-BDE+\left(B^2-4AC\right)F\right)}{\left(4AC-B^2\right)^\frac32}$

This is where we can look at what happens when the conic is not an ellipse. Note that for the denominator to be real and non-zero, we need $4AC>B^2$ ; but this is exactly the condition for the conic to be an ellipse. This makes intuitive sense; if a conic encloses an area, it must be an ellipse.

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Putting everything together and tidying up, we have $T=\frac{31\pi \left(84B^2+8B-1\right)}{2 \left(-B^2 + 16 B - 2\right)^\frac32}$

Setting the derivative equal to zero gives the equation $84 B^3 + 688 B^2-403 B + 8 = 0$

Solving this numerically, the minimum area is found when $B\approx 0.52956$ , giving area $T\approx 84.664$ and answer $\boxed{8466}$ .