The curve represented by $\frac{x^2}{12-\alpha}+\frac{y^2}{\alpha-10}=1$ is

Let $x$ be the eccentricity of a hyperbola whose transverse axis is twice its conjugate axis. Let $y$ be the eccentricity of another hyperbola for which the distance between the focii is 3 times the distance between its directrices. Then $y^2-x^2=$

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is $\frac{36}{13}$ and its eccentricity is $\frac{\sqrt{13}}{3}$, then $a-b=$