Standard Equation of a Hyperbola

Deriving the standard equation $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$.

The derivation runs in exact parallel to the ellipse's (section 7.2.1), with $e>1$ in place of $0<e<1$. Let $S$ be the focus, $d$ the directrix, $e$ the eccentricity. Draw $SZ\perp d$; let $A,A'$ divide $SZ$ internally and externally in the ratio $e:1$ (both lie on the hyperbola by definition). Let $AA'=2a$, with midpoint $O$ as origin: $A(a,0),\,A'(-a,0)$.

Exactly the same section-formula algebra as for the ellipse gives $S\equiv(ae,0)$ and directrix $x=\dfrac{a}{e}$.

For $P(x,y)$ on the hyperbola: $SP=\sqrt{(x-ae)^2+y^2}$, $PM=\left|x-\dfrac{a}{e}\right|$. Substituting $SP=e\cdot PM$ and squaring:

Here is the KEY difference from the ellipse: since $e>1$, we have $1-e^2<0$. Multiplying both sides by $-1$:

Dividing by $a^2(e^2-1)$ (a positive quantity, since $e>1$):

Writing $b^2=a^2(e^2-1)$, this becomes the standard equation of a hyperbola:

Notice the relation is $b^2=a^2(e^2-1)$ (a MINUS one, since $e>1$) — the mirror image of the ellipse's $b^2=a^2(1-e^2)$. As with the ellipse, a second focus $S'(-ae,0)$ and second directrix $x=-a/e$ exist symmetrically, and it can be verified that $PS'=e\cdot PM'$ for the same point $P$.

Worked Example 1 — full property set for two given hyperbolas.

1. $\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$: $a^2=4,b^2=12\Rightarrow a=2,b=2\sqrt3$; transverse axis $=4$, conjugate axis $=4\sqrt3$; $e=\sqrt{\dfrac{a^2+b^2}{a^2}}=\sqrt{\dfrac{16}{4}}=2$; $ae=4$, foci $(\pm4,0)$; $a/e=1$, directrices $x=\pm1$; latus rectum $=\dfrac{2(12)}{2}=12$.
2. $\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$ (transverse axis along $Y$): here $a^2=16,b^2=9\Rightarrow a=4,b=3$ (using the roles $A^2$ under $y^2$, but following the book's own labelling of $a,b$ for this sub-case); transverse axis $=6$, conjugate axis $=8$; $e=\sqrt{\dfrac{a^2+b^2}{a^2}}=\dfrac53$; foci $(0,\pm5)$; directrices $y=\pm\dfrac95$; latus rectum $=\dfrac{2(16)}{3}=\dfrac{32}{3}$. …