The hyperbola

What defines a hyperbola

• A hyperbola is the set of points whose distances to two fixed points (the foci) DIFFER by a constant.
• Compare with the ellipse: the ellipse adds the two distances, the hyperbola subtracts them. That one change turns a closed oval into two separate open branches.
• Practical uses: long-range navigation systems locate a ship by the difference in arrival times of two signals, which places it on a hyperbola; cooling towers are hyperbolic in profile; a comet with too much energy to be captured follows a hyperbolic path.

The standard equation

$$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \qquad \text{opens LEFT and RIGHT}$$

$$\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1 \qquad \text{opens UP and DOWN}$$

• The minus sign identifies the hyperbola.
• The POSITIVE term tells you which way it opens — not the size of the denominators. This is the key difference from the ellipse, where the larger denominator decides.
  • $x$ term positive → branches open left and right, vertices on a horizontal line
  • $y$ term positive → branches open up and down
• $(h, k)$ is the centre, which is not on the curve — it is the midpoint between the two branches.

Vertices and asymptotes

• The vertices are the two points where the branches are closest together, at distance $a$ from the centre along the opening direction (for the $x$-positive form): $(h \pm a,\ k)$.
• The asymptotes are the two straight lines the branches approach but never touch:

$$y - k = \pm\frac{b}{a}(x - h) \qquad \text{for the } x\text{-positive form}$$

• How to find them reliably: replace the 1 on the right-hand side with 0 and solve:

$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 0 \quad \Rightarrow \quad \frac{(y-k)}{b} = \pm\frac{(x-h)}{a}$$

• This works for both orientations and removes any need to memorise which of $a$ and $b$ goes on top.

• Sketching method — draw the box. Mark $\pm a$ horizontally and $\pm b$ vertically from the centre, draw the rectangle, extend its diagonals as the asymptotes, then draw each branch through its vertex hugging the asymptotes.

The foci

$$c^2 = a^2 + b^2$$

• Note the ADDITION. For an ellipse it was a subtraction; for a hyperbola the foci lie beyond the vertices, so $c > a$ always.
• The foci sit on the same axis as the vertices, at $(h \pm c,\ k)$ for the $x$-positive form.
• Eccentricity is again $e = \dfrac{c}{a}$, but now $e > 1$ for every hyperbola.

The rectangular hyperbola

• When $a = b$ the asymptotes are perpendicular, and the hyperbola is called rectangular.
• Rotated to sit against the axes, the rectangular hyperbola has the familiar form

$$xy = c^2$$

• Its asymptotes are the coordinate axes themselves.
• This is the shape of an inverse relationship — Boyle's law $PV = k$, and any "constant product" context.
• The vertices of $xy = c^2$ are at $(c, c)$ and $(-c, -c)$, on the line $y = x$.

Completing the square for a hyperbola

• The general form has both squared terms with OPPOSITE signs:

$$16x^2 - 9y^2 - 64x - 54y - 161 = 0$$

• The procedure is the same as for an ellipse, but watch the negative factor:
  • $$16(x^2 - 4x) - 9(y^2 + 6y) = 161$$
  • $$16(x^2 - 4x + 4) - 9(y^2 + 6y + 9) = 161 + 16(4) - 9(9)$$
  • $$16(x-2)^2 - 9(y+3)^2 = 161 + 64 - 81 = 144$$
  • Divide by 144: $\dfrac{(x-2)^2}{9} - \dfrac{(y+3)^2}{16} = 1$
  • Centre $(2, -3)$, $a = 3$, $b = 4$, opening left and right; asymptotes $y + 3 = \pm\frac{4}{3}(x - 2)$
• Adding 9 inside a bracket multiplied by $-9$ SUBTRACTS 81 from the equation. Getting that sign wrong is the most common error on this page.