Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV resit / 2

A hyperbola with focus of a conic $F$, directrix of a conic $D$ not containing $F$, and eccentricity $e>1$ is the locusPut $c=\sqrt{a^2+b^2}$ and $e=c/a>1$. For $F=(c,0)$ and $D=\{x=a^2/c\}$, the defining equality becomesBoth sides are nonnegative, so squaring is reversible. Expanding and cancelling the linear terms givesThus the two matched focus of a conic and directrix of a conic choices, related by reflection, areThe signs in each pair must agree.

On the hyperbola, $y=\pm (b/a)\sqrt{x^2-a^2}$. As $|x|\to \infty$,Hence the perpendicular distance to one of the two straight lines tends to zero on each end of a branch. The asymptotes are therefore

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