There are two forms of rectangular hyperbola $x^2-y^2 = a^2$ and xy = $c^2$. How do we find the focus, latus Rectum and Directrix of xy = $c^2$. For the former there is a set of well defined formulas.

  • $\begingroup$ Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$ $\endgroup$ Aug 30, 2017 at 2:40
  • $\begingroup$ For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$. $\endgroup$
    – Blue
    Aug 30, 2017 at 3:23
  • $\begingroup$ Rotate the formulas. $\endgroup$
    – amd
    Aug 30, 2017 at 4:47

1 Answer 1


You can find what you need without rotations, just remembering the geometric properties of foci $F$ and $F'$. First of all, they lie on the transverse (or major) axis of the hyperbola (line $y=x$ in our case) and are symmetric with respect to center $(0,0)$. We can then conveniently set $F=(f,f)$ and $F'=(-f,-f)$, with $f>0$.

The hyperbola is the locus of points $P$ such that $$|F'P-FP|=k,$$ where $k$ is a constant. $A=(c,c)$ and $B=(-c,-c)$ are the vertices of the hyperbola (that is its intersections with the transverse axis): as $|F'A-FA|=AB=2\sqrt2c$ we then get $k=2\sqrt2c$. Apply now the equation to any convenient point on the hyperbola, for instance $P=(f,c^2/f)$: you have $FA=f-c^2/f$, $F'A=\sqrt{4f^2+(f+c^2/f)^2}$ and from the above equation, after some algebra, you can get $f=\sqrt2c$.

To find the latus rectum $FL$ you just need to find $L$ as one of the intersections between the hyperbola and a line through $F$ perpendicular to the transverse axis. Similarly, from the definition of directrix you can find its equation.


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