Hyperbolas
Hyperbolas
The set of all points whose distances from two foci have a constant difference of $2a$.
Horizontal transverse axis: $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
Vertical transverse axis: $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$
$$c^2 = a^2 + b^2$$
Asymptotes (centered at origin, horizontal): $y = \pm \frac{b}{a}x$
Asymptotes (vertical): $y = \pm \frac{a}{b}x$
Find the vertices, foci, and asymptotes: $\frac{x^2}{9}-\frac{y^2}{16}=1$.
$a=3$, $b=4$, $c = \sqrt{9+16} = 5$.
Vertices: $(\pm 3, 0)$. Foci: $(\pm 5, 0)$. Asymptotes: $y = \pm \frac{4}{3}x$.
Write the equation: vertices $(0, \pm 4)$, foci $(0, \pm 5)$.
Vertical transverse. $a = 4$, $c = 5$, $b^2 = 25 - 16 = 9$.
$$\frac{y^2}{16} - \frac{x^2}{9} = 1$$
Find the asymptotes: $\frac{y^2}{25} - \frac{x^2}{4} = 1$.
Vertical transverse: $y = \pm \frac{a}{b}x = \pm \frac{5}{2}x$.
Practice Problems
Show Answer Key
1. $a=2$, vertices $(\pm 2,0)$; $c=4$, foci $(\pm 4,0)$
2. $\frac{x^2}{36}-\frac{y^2}{64}=1$
3. $y = \pm \frac{3}{7}x$
4. Vertical
5. $c = \sqrt{25+144} = 13$
6. $b^2 = 100-36 = 64$; $\frac{x^2}{36}-\frac{y^2}{64}=1$
7. Hyperbola: $c^2 = a^2+b^2$; Ellipse: $c^2 = a^2-b^2$
8. Center $(-1,3)$, $a = 5$
9. $b/a = 2$, so $b = 6$
10. $e = c/a > 1$
11. $2b = 6$
12. Hyperbola (divide by 36: $\frac{x^2}{9}-\frac{y^2}{4}=1$)