Let \((x_1,y_1)\) and \((x_2,y_2)\) be 2 points in the plane. Suppose we would like to find ellipses through both points that have radius \(a>0\) on the \(x\) axis and radius \(b>0\) on the \(y\) axis. We are looking for points \((h,k)\) that satisfy the following:
\[\begin{align}& {(x_1-h)^2\over a^2}+{(y_1-k)^2\over b^2}=1\\& {(x_2-h)^2\over a^2}+{(y_2-k)^2\over b^2}=1 \end{align}\]This is simplified considerably when either \(x_1=x_2\) or \(y_1=y_2\). The general case involves a lot of gross algebra. For now, suppose both \(x_1=x_2\) and \(y_1=y_2\). Then both of the above equations are identical and we only need to solve one. By choosing \(h\) or \(k\), one of the terms becomes a constant and the other variable is found by solving a quadratic equation. There are infinitely many such ellipses with centers \((h,k)\) that are on the ellipse with radii \(a\) in the \(x\) axis and \(b\) in the \(y\) axis with center \((x_1,y_1)\).
Now if just one of \(x_1=x_2\) or \(y_1=y_2\), then the 2 points are along an axis. One of the terms is identical in both equations so it can be canceled and makes the algebra a lot easier. First, suppose \(x_1=x_2\). Then we have \((y_1-k)^2=(y_2-k)^2\) which implies \(k=(y_1+y_2)/2\). By substituting this in, we find \(h=x_1\pm{a\over b}\sqrt{b^2-(y_1-y_2)^2/4}\). Similarly, for \(y_1=y_2\), we find \(h=(x_1+x_2)/2\) and \(k=y_1\pm{b\over a}\sqrt{a^2-(x_1-x_2)^2/4}\).
Things are considerably more complicated when \(x_1\neq x_2\) and \(y_1\neq y_2\). By subtracting the 2 eqquations, expanding the squares, and simplifying, we get:
\[b^2(x_1^2-x_2)^2+a^2(y_1^2-y_2^2)=2hb^2(x_1-x_2)+2ka^2(y_1-y_2)\]Let this be written as \(c=c_1h+c_2k\) with the constants \(c=b^2(x_1^2-x_2^2)+a^2(y_1^2-y_2^2)\), \(c_1=2b^2(x_1-x_2)\), \(c_2=2a^2(y_1-y_2)\). Writing the solution in terms of the given values gets really disgusting. I got lazy and did not feel like finishing the algebra right now. From here, we can use the linear relationship \(c=c_1h+c_2k\) to turn one of the equations into a single variable quadratic equation. After solving that, use this linear relationship to find the value of the other variable.