Consider an ellipse $\frac{x^2}{36}+\frac{y^2}{18}=1$
There is a hyperbola whose one asymptote is the major axis of a given ellipse. If the eccentricity of the given ellipse and hyperbola are reciprocal to each other, both have the same center and both touch each other in the first and third quadrant. The focus of hyperbola is at?
Clearly, it will be a rectangular hyperbola say $xy=k$ for $k>0$, and if hyperbola and ellipse touch each other we get biquadratic equation $x^4-36x^2+2k^2=0$. I can consider $D=0$ for tangency, but as it is quadratic in $x^2$ it may have another condition when one root is positive and another root is negative. That's where I am stuck. Please help!
E: $$x^2/36+y^2/18=1 \implies y'=-\frac{x}{2y}$$ RK: $$xy=k^2 \implies y'=-y/x$$ Equating the two derivatives (slopes), we get $2y^2=x^2$ Putting this in RH we get $x= 2^{1/4}k, y= 2^{-1/4} k$ in the first quadrant. Using these in E we get $\frac{k\sqrt{2}}{36}+\frac{k}{18\sqrt{2}}=1 \implies k=3 (2)^{1/4}$. So the required Two RH are $$xy= \pm 9 \sqrt{2}.$$