The Hyperbola

Definition and the Standard Equation

Key Point (Definition): A hyperbola is the set of all points whose distances from two fixed points (the foci) have a constant difference — farther distance minus closer distance.

The midpoint of the foci is the centre; the line through the foci is the transverse axis; the perpendicular line through the centre is the conjugate axis; the intersections with the transverse axis are the vertices. Write $2c$ for the focal distance, $2a$ for the distance between vertices, and define

$$b = \sqrt{c^2 - a^2}, \qquad\text{i.e.}\qquad c^2 = a^2 + b^2$$

Taking the point $P$ at a vertex shows the constant difference is exactly $2a$.

Deriving the equation. With foci $(\pm c, 0)$ and $PF_1 - PF_2 = 2a$, two squarings give

$$\frac{x^2}{a^2} - \frac{y^2}{c^2 - a^2} = 1 \;\Rightarrow\; \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Since $\frac{x^2}{a^2} = 1 + \frac{y^2}{b^2} \ge 1$, no part of the curve lies between the lines $x = -a$ and $x = a$: the hyperbola splits into two branches.

Key Point (Eccentricity): $e = \frac{c}{a} \ge 1$ — never less than one, because $c \ge a$. The foci sit at distance $ae$ from the centre. A hyperbola with $a = b$ is called equilateral (its eccentricity is $\sqrt{2}$).

Both Orientations, Latus Rectum, and the Workflow

Key Point (Standard equations): Transverse axis on the $x$-axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, foci $(\pm c, 0)$. Transverse axis on the $y$-axis: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, foci $(0, \pm c)$. In both, $c^2 = a^2 + b^2$.

Which is which? Follow the positive term: its denominator is $a^2$ and its variable names the transverse axis. Note carefully — unlike the ellipse, it is not about which denominator is larger.

Key Point (Latus rectum): as for the ellipse, the focal chord perpendicular to the transverse axis has length $\frac{2b^2}{a}$.

The standard workflow:

Equation → data: for $\frac{x^2}{9} - \frac{y^2}{16} = 1$: $a = 3$, $b = 4$, $c = \sqrt{9 + 16} = 5$; foci $(\pm 5, 0)$, vertices $(\pm 3, 0)$, $e = \frac{5}{3}$, latus rectum $\frac{32}{3}$. For $y^2 - 16x^2 = 16$, divide by 16 first: $\frac{y^2}{16} - \frac{x^2}{1} = 1$.

Data → equation: foci $(0, \pm 3)$ and vertices $\left(0, \pm\frac{\sqrt{11}}{2}\right)$ give $a^2 = \frac{11}{4}$, $b^2 = 9 - \frac{11}{4} = \frac{25}{4}$: $100y^2 - 44x^2 = 275$. Foci $(0, \pm 12)$ with latus rectum 36 give $b^2 = 18a$ and $144 = a^2 + 18a$, so $a = 6$, $b^2 = 108$: $3y^2 - x^2 = 108$.

[Board Tip] For the hyperbola the sum $c^2 = a^2 + b^2$ means $c$ is the biggest of the three — the foci always sit beyond the vertices. If your computed focus lands between the vertices, you have used the ellipse relation by mistake.

Solved Examples

Example 1: Both orientations

Find the foci, vertices, eccentricity and latus rectum of: (i) $\frac{x^2}{9} - \frac{y^2}{16} = 1$; (ii) $y^2 - 16x^2 = 16$.

Solution:

Step 1 — (i) Identify. Positive term in $x$: transverse on the $x$-axis, $a = 3$, $b = 4$.

Step 2 — (i) Compute $c$ and list. $c = \sqrt{9 + 16} = 5$: foci $(\pm 5, 0)$, vertices $(\pm 3, 0)$, $e = \frac{5}{3}$, latus rectum $\frac{2(16)}{3} = \frac{32}{3}$.

Step 3 — (ii) Normalise. $\frac{y^2}{16} - \frac{x^2}{1} = 1$: transverse on the $y$-axis, $a = 4$, $b = 1$, $c = \sqrt{17}$.

Step 4 — (ii) List. Foci $(0, \pm\sqrt{17})$, vertices $(0, \pm 4)$, $e = \frac{\sqrt{17}}{4}$, latus rectum $\frac{2(1)}{4} = \frac{1}{2}$.

Takeaway: ADD to get $c$ for a hyperbola — the foci must land beyond the vertices, a built-in sanity check.

Example 2: Reading practice

Analyse: (i) $\frac{x^2}{16} - \frac{y^2}{9} = 1$; (ii) $\frac{y^2}{9} - \frac{x^2}{27} = 1$; (iii) $9y^2 - 4x^2 = 36$.

Solution:

Step 1 — (i). $a = 4$, $b = 3$, $c = 5$: foci $(\pm 5, 0)$, vertices $(\pm 4, 0)$, $e = \frac{5}{4}$, latus rectum $\frac{9}{2}$.

Step 2 — (ii). $a = 3$, $b^2 = 27$, $c = \sqrt{36} = 6$: foci $(0, \pm 6)$, vertices $(0, \pm 3)$, $e = 2$, latus rectum $\frac{2(27)}{3} = 18$.

Step 3 — (iii). Normalise: $\frac{y^2}{4} - \frac{x^2}{9} = 1$: $a = 2$, $b = 3$, $c = \sqrt{13}$: foci $(0, \pm\sqrt{13})$, vertices $(0, \pm 2)$, $e = \frac{\sqrt{13}}{2}$, latus rectum $\frac{2(9)}{2} = 9$.

Takeaway: The transverse axis follows the POSITIVE term, not the larger denominator — the hyperbola's orientation rule differs from the ellipse's.

Example 3: Rescale first

Analyse: (i) $16x^2 - 9y^2 = 576$; (ii) $5y^2 - 9x^2 = 36$; (iii) $49y^2 - 16x^2 = 784$.

Solution:

Step 1 — (i). Divide by 576: $\frac{x^2}{36} - \frac{y^2}{64} = 1$: $a = 6$, $b = 8$, $c = 10$: foci $(\pm 10, 0)$, vertices $(\pm 6, 0)$, $e = \frac{5}{3}$, latus rectum $\frac{2(64)}{6} = \frac{64}{3}$.

Step 2 — (ii). $\frac{y^2}{36/5} - \frac{x^2}{4} = 1$: $a^2 = \frac{36}{5}$, $b^2 = 4$, $c^2 = \frac{56}{5}$: foci $\left(0, \pm\frac{2\sqrt{70}}{5}\right)$, $e = \sqrt{\frac{56/5}{36/5}} = \frac{\sqrt{14}}{3}$, latus rectum $\frac{2(4)}{6/\sqrt{5}} = \frac{4\sqrt{5}}{3}$.

Step 3 — (iii). $\frac{y^2}{16} - \frac{x^2}{49} = 1$: $c = \sqrt{65}$: foci $(0, \pm\sqrt{65})$, vertices $(0, \pm 4)$, $e = \frac{\sqrt{65}}{4}$, latus rectum $\frac{2(49)}{4} = \frac{49}{2}$.

Takeaway: Fractional $a^2$ values are fine — carry them exactly and simplify only at the end.

Example 4: From vertices and foci

Find the hyperbola with: (i) foci $(0, \pm 3)$, vertices $\left(0, \pm\frac{\sqrt{11}}{2}\right)$; (ii) vertices $(\pm 2, 0)$, foci $(\pm 3, 0)$; (iii) vertices $(0, \pm 5)$, foci $(0, \pm 8)$; (iv) vertices $(0, \pm 3)$, foci $(0, \pm 5)$.

Solution:

Step 1 — (i). $a^2 = \frac{11}{4}$, $c^2 = 9$: $b^2 = 9 - \frac{11}{4} = \frac{25}{4}$: $\frac{y^2}{11/4} - \frac{x^2}{25/4} = 1$, i.e. $100y^2 - 44x^2 = 275$.

Step 2 — (ii). $b^2 = 9 - 4 = 5$: $\frac{x^2}{4} - \frac{y^2}{5} = 1$.

Step 3 — (iii). $b^2 = 64 - 25 = 39$: $\frac{y^2}{25} - \frac{x^2}{39} = 1$.

Step 4 — (iv). $b^2 = 25 - 9 = 16$: $\frac{y^2}{9} - \frac{x^2}{16} = 1$.

Takeaway: $b^2 = c^2 - a^2$ every time — vertices give $a$, foci give $c$, subtraction gives the missing piece.

Example 5: From an axis length

Find the hyperbola with: (i) foci $(\pm 5, 0)$ and transverse axis of length 8; (ii) foci $(0, \pm 13)$ and conjugate axis of length 24.

Solution:

Step 1 — (i). $2a = 8$: $a = 4$, $c = 5$: $b^2 = 25 - 16 = 9$: $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Step 2 — (ii). Conjugate axis 24 gives $b = 12$; $c = 13$: $a^2 = 169 - 144 = 25$: $\frac{y^2}{25} - \frac{x^2}{144} = 1$.

Takeaway: Transverse length fixes $a$; conjugate length fixes $b$ — read which one the problem hands you.

Example 6: Latus rectum data

Find the hyperbola with: (i) foci $(0, \pm 12)$, latus rectum 36; (ii) foci $(\pm 3\sqrt{5}, 0)$, latus rectum 8; (iii) foci $(\pm 4, 0)$, latus rectum 12.

Solution:

Step 1 — (i) Set up the quadratic. $\frac{2b^2}{a} = 36$ gives $b^2 = 18a$; with $c^2 = a^2 + b^2$: $144 = a^2 + 18a$.

Step 2 — (i) Solve. $a^2 + 18a - 144 = 0$: $(a + 24)(a - 6) = 0$: $a = 6$, $b^2 = 108$: $\frac{y^2}{36} - \frac{x^2}{108} = 1$, i.e. $3y^2 - x^2 = 108$.

Step 3 — (ii). $b^2 = 4a$, $45 = a^2 + 4a$: $(a+9)(a-5) = 0$: $a = 5$, $b^2 = 20$: $\frac{x^2}{25} - \frac{y^2}{20} = 1$.

Step 4 — (iii). $b^2 = 6a$, $16 = a^2 + 6a$: $(a+8)(a-2) = 0$: $a = 2$, $b^2 = 12$: $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

Takeaway: Latus rectum + foci always produces a quadratic in $a$ — the negative root discards itself.

Example 7: From eccentricity

Find the hyperbola with vertices $(\pm 7, 0)$ and $e = \frac{4}{3}$.

Solution:

Step 1 — Find $c$. $a = 7$: $c = ae = \frac{28}{3}$.

Step 2 — Find $b^2$. $b^2 = c^2 - a^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}$.

Step 3 — Write. $\frac{x^2}{49} - \frac{9y^2}{343} = 1$.

Takeaway: $e$ converts $a$ into $c$ in one multiplication; the relation then yields $b^2$ as usual.

Example 8: Through a point

Find the hyperbola with foci $(0, \pm\sqrt{10})$ passing through $(2, 3)$.

Solution:

Step 1 — Set up with one unknown. $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ with $b^2 = 10 - a^2$: substituting $(2,3)$: $\frac{9}{a^2} - \frac{4}{10 - a^2} = 1$.

Step 2 — Clear denominators. $9(10 - a^2) - 4a^2 = a^2(10 - a^2)$ gives $a^4 - 23a^2 + 90 = 0$.

Step 3 — Solve and audit. $(a^2 - 5)(a^2 - 18) = 0$: $a^2 = 5$ or 18; but $a^2 < c^2 = 10$ kills 18.

Step 4 — Write. $a^2 = b^2 = 5$: $\frac{y^2}{5} - \frac{x^2}{5} = 1$ — an equilateral hyperbola.

Takeaway: Always audit quadratic-in-$a^2$ roots against $a^2 < c^2$ — one root is usually geometric nonsense.

Example 9: Eccentricity of the equilateral hyperbola

Show that every hyperbola with $a = b$ has eccentricity $\sqrt{2}$.

Solution:

Step 1 — Apply the relation. $c^2 = a^2 + b^2 = 2a^2$.

Step 2 — Take the ratio. $c = a\sqrt{2}$, so $e = \frac{c}{a} = \sqrt{2}$ — independent of $a$. ∎

Takeaway: All equilateral hyperbolas are similar — one shape at every scale, like all circles.