Change one word in the definition of an ellipse: instead of the sum of the distances to two foci, keep the difference constant. The curve splits into two branches that open away from each other, and that is a hyperbola.
Key idea
A hyperbola is the set of points whose distances to two foci differ by a constant, $2a$. $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ opens left and right, with vertices $(\pm a, 0)$, foci $(\pm c, 0)$ where $c^2 = a^2 + b^2$, and asymptotes $y = \pm\frac{b}{a}x$. $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ opens up and down, with vertices $(0, \pm a)$, foci $(0, \pm c)$ and asymptotes $y = \pm\frac{a}{b}x$.
Why it works
The branches open along the axis of the positive squared term: in $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, putting $x = 0$ would need $-\frac{y^2}{b^2} = 1$, which is impossible, so the curve never crosses the $y$-axis. The foci lie beyond the vertices, so $c > a$, and $b$ is defined by $c^2 = a^2 + b^2$. Far from the center the $1$ hardly matters, so the curve hugs $\frac{x^2}{a^2} = \frac{y^2}{b^2}$, the lines $y = \pm\frac{b}{a}x$.
Worked examples
Worked example: All the features
Describe $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.
- The $x^2$ term is positive: it opens left and right, with $a = 3$, $b = 4$.
- $c^2 = 9 + 16 = 25$, $c = 5$.
- Asymptotes $y = \pm\frac{4}{3}x$.
Answer: Vertices $(\pm 3, 0)$, foci $(\pm 5, 0)$, asymptotes $y = \pm\frac{4}{3}x$
Worked example: Opening up and down
Describe $\dfrac{y^2}{4} - \dfrac{x^2}{9} = 1$.
- The $y^2$ term is positive: it opens up and down, with $a = 2$ (under $y^2$) and $b = 3$.
- $c = \sqrt{4 + 9} = \sqrt{13}$.
- Asymptotes $y = \pm\frac{a}{b}x = \pm\frac{2}{3}x$.
Answer: Vertices $(0, \pm 2)$, foci $\left(0, \pm\sqrt{13}\right)$, asymptotes $y = \pm\frac{2}{3}x$
Worked example: Writing an equation
Write the hyperbola with vertices $(0, \pm 6)$ and foci $(0, \pm 10)$.
- $a = 6$, $c = 10$, so $b^2 = c^2 - a^2 = 64$.
- The vertices are on the $y$-axis, so the $y^2$ term is positive.
Answer: $\frac{y^2}{36} - \frac{x^2}{64} = 1$
Common mistake
Drawing the branches the wrong way, or borrowing the ellipse rule $c^2 = a^2 - b^2$. Checks: the branches open toward the variable with the plus sign; and the foci are outside the vertices, so $c$ must be bigger than $a$.