An ellipse touches the $x$ axis. The length of the major axis is $2a$ while the minor axis is given as $2b$, What would be the locus of its focus? ($a>b$.)
My Approach:
If the foci are $(x_1,y_1$ and $(x_2,y_2)$, then $(x_1-x_2)^2 + (y_1 - y_2)^2 = 4(a^2 - b^2),$ also $y_1y_2 = b^2,$ since product of perpendicular subtended from the foci on any tangent is $b^2$.
These are the properties which seem relevant in this question but I am not able to use them to find out the answer. I also might be missing out on other things.
Any hints/solutions to the problem are appreciated.
HINT.
The ellipse touches the $x$ axis at the origin (see comments to the question). Hence, if $F_1=(x,y)$, then $\displaystyle F_2=\left(-{b^2\over y^2}x, {b^2\over y}\right)$.
Insert these into $\overline{F_1F_2}^2=4(a^2-b^2)$ to get the desired relation.