Q.Find the equation of the hyperbola whose vertices are $(0,\pm6)$ and eccentricity is $\dfrac{5}{3}$.

A hyperbola is the locus of a point whose distances to two fixed foci differ, in absolute value, by a constant $2a$ (with $2a$ less than the distance between the foci, the reverse inequality from the ellipse); this gives the standard equation $x^2/a^2-y^2/b^2=1$, where $b^2=c^2-a^2$, or equivalently $b^2=a^2(e^2-1)$ in terms of the eccentricity $e=c/a$, which always exceeds $1$ for a genuine hyperbola -- again the opposite range from the ellipse. The foci sit at $(\pm c,0)=(\pm ae,0)$ and the vertices at $(\pm a,0)$, with transverse axis length $2a$, conjugate axis length $2b$, and latus-rectum length $2b^2/a$, the same formula as the ellipse's. Uniquely among the four …