Q.Show that the absolute value of difference of the focal distances of any point $P$ on the hyperbola is the length of its transverse axis.

Matching $xy=16$ to the standard rectangular hyperbola $xy=c^2$ gives $c=4$, so the vertices are $(4,4)$ and $(-4,-4)$.

1. A rectangular hyperbola referred to its asymptotes as axes has equation $xy=c^2$, with asymptotes being the coordinate axes and axis of symmetry the line $y=x$.
2. Its vertices are the points where the curve meets the line $y=x$: substitute $y=x$ into $xy=c^2$ to get $x^2=c^2$, so $x=\pm c$, giving vertices $(c,c)$ and $(-c,-c)$. …

For $xy=c^2=18$, $c=3\sqrt2$, and the foci of the rectangular hyperbola (referred to its asymptotes as axes) are at $(\pm c\sqrt2,\pm c\sqrt2)=(\pm6,\pm6)$.

1. Write $xy=18$ in the standard form $xy=c^2$, so $c^2=18$, $c=\sqrt{18}=3\sqrt2$. …

Put the given equation in standard hyperbola form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then use the standard asymptote formula $y=\pm\frac{b}{a}x$.

1. Given: $36y^2-25x^2+900=0$.
2. Rearrange: $-25x^2+36y^2=-900 \Rightarrow 25x^2-36y^2=900$.
3. Divide by 900: $\dfrac{25x^2}{900}-\dfrac{36y^2}{900}=1 \Rightarrow \dfrac{x^2}{36}-\dfrac{y^2}{25}=1$. …

The directrices of the given hyperbola are $x=\pm\dfrac{8}{\sqrt5}$.

1. Divide $x^2-4(y-3)^2=16$ by $16$: $\dfrac{x^2}{16}-\dfrac{(y-3)^2}{4}=1$.
2. This is centred at $(0,3)$ with transverse axis horizontal, $a^2=16\Rightarrow a=4$, $b^2=4\Rightarrow b=2$.
3. Eccentricity: $e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac14}=\sqrt{\dfrac54}=\dfrac{\sqrt5}{2}$.
4. Directrices (measured from the centre, since the transverse axis is horizontal): $x=h\pm\dfrac{a}{e}=0\pm\dfrac{4}{\sqrt5/2}=\pm\dfrac{8}{\sqrt5}$. …