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For two distinct intersecting straight lines through the origin, with independent real linear equations $L_1=0$ and $L_2=0$, the nondegenerate centred hyperbolas having exactly those asymptotes areIndeed, the homogeneous quadratic form of a centred hyperbola vanishes in its two asymptotic directions and hence is a nonzero scalar multiple of $L_1{L}_2$. Its equation has no linear terms because the origin is its centre, and its constant term is nonzero by nondegeneracy. Conversely, the invertible linear transformation $(x,y)\mapsto (L_1,L_2)$ takes the displayed curve to $uv=c$, a hyperbola, and its ends approach $u=0$ or $v=0$. Distinct values of $c$ give disjoint level sets. In particular, the rectangular hyperbolas with asymptotes $y=\pm x$ are $x^2-y^2=c$, $c\ne 0$.

A nondegenerate hyperbola centred at the origin has a quadratic equationThere are no linear terms because the origin is its centre. Its asymptotes are the zero set of the homogeneous quadratic form $A{x}^2+2Bxy+C{y}^2$. Requiring the directions $y=x$ and $y=-x$ gives $A+2B+C=A-2B+C=0$, hence $B=0$ and $C=-A$. Since the conic section is nondegenerate, $A$ and $F$ are nonzero. Thus all such hyperbolas with prescribed asymptotes are exactlyFor $c>0$ these rectangular hyperbolas open along the $x$-axis, and for $c<0$ along the $y$-axis. Conversely, each displayed hyperbola has the required centre and asymptotes. Two distinct members have different values of $x^2-y^2$, so they cannot intersect.

For the second pair of asymptotes, the same argument gives $H_2:4x^2-y^2=d$, $d\ne 0$. The intersections of concentric hyperbolas therefore satisfyIf both right sides are positive, there are four independent sign choices. If precisely one is zero and the other is positive, there are two points. If either is negative there are none; both cannot vanish because $c\ne 0$. Each possibility occurs: take $(c,d)=(1,5)$ for four, $(1,4)$ for two, and $(1,1)$ for none. Hence