The rectangular hyperbolaEdexcel International A Level Further Maths: Revision notes

A rectangular hyperbola has Cartesian equation $xy=c^2,$ where $c$ is a positive constant. It has two separate branches, one in the first quadrant ($x,y>0$) and one in the third ($x,y<0$), and the coordinate axes are its asymptotes: the curve gets closer and closer to them but never touches them. The product of the coordinates of every point on it is the same constant $c^2$. Example: $xy=36$ has $c=6$. The point $(4,9)$ lies on it because $4\times9=36$, but $(5,6)$ does not because $5\times6=30$.

Every point on $xy=c^2$ can be written using one parameter $t$: $x=ct,\qquad y=\frac ct,\qquad t\neq0.$ Multiplying gives $xy=ct\times\frac ct=c^2$, so these equations always satisfy the Cartesian equation. Positive values of $t$ give the first-quadrant branch and negative values the third-quadrant branch. To go from parametric to Cartesian, multiply $x$ by $y$. To go the other way, read off $c$ from $c^2$ and use $x=ct$, $y=\frac ct$. For $xy=36$: $x=6t$, $y=\frac6t$.

The point $P\left(ct,\frac ct\right)$ is the general point on $xy=c^2$: choosing a value of $t$ picks out one particular point.

• Given $t$: substitute. For $xy=36$ and $t=3$ the point is $(18,2)$.
• Given a coordinate: use $x=ct$ to find $t$, then find $y$. For $xy=36$ and $x=-9$, $t=-\frac32$ and $y=\frac{6}{-3/2}=-4$.
• Changing $t$ to $\frac1t$ swaps the coordinates, reflecting the point in the line $y=x$.
• Different values of $t$ always give different points.

Because the general point has only one unknown, $t$, problems about points on the curve reduce to algebra in $t$.

• Gradient of a chord: for points with parameters $p$ and $q$ on $x=ct,\ y=\frac ct$, $\frac{\frac cq-\frac cp}{cq-cp}=\frac{c(p-q)/pq}{c(q-p)}=-\frac1{pq}.$
• Distance from the origin: $OP^2=c^2t^2+\frac{c^2}{t^2}$.
• Area under the corners: the rectangle from $O$ to $P$ has area $ct\times\frac ct=c^2$, the same for every $t>0$.
• Meeting a line: substitute $x=ct$, $y=\frac ct$ into the line equation and solve for $t$. For $xy=25$ meeting $y=x$: $5t=\frac5t$, so $t=\pm1$ and the points are $(5,5)$ and $(-5,-5)$.

• Given $xy=c^2$ and asked whether a point lies on the curve: the Cartesian form is quicker.
• Asked for a general point, a chord, distances or midpoints: use the parametric form.
• Always check $t\neq0$, and that your value of $t$ has the same sign as $x$ (since $x=ct$ with $c>0$).