The Equation of a Hyperbola

One sign changes

The ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$x²/9 + y²/4 = 1 is a closed curve around the origin: it reaches 3 along the x-axis and 2 along the y-axis. Keep the same numbers and change the plus sign to a minus: $\frac{x^2}{9}-\frac{y^2}{4}=1$x²/9 − y²/4 = 1. This curve is a hyperbola, and it is not closed. It comes in two separate pieces, called its branches.

Where it meets the x-axis

Put y = 0: $\frac{x^2}{9}=1$x²/9 = 1, so $x^2=9$x² = 9 and x = 3 or x = −3. The hyperbola meets the x-axis at (3, 0) and (−3, 0). These two points are its vertices, where each branch turns back. The origin, halfway between them, is its center.

The hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$x²/9 − y²/4 = 1. Its two branches turn at (−3, 0) and (3, 0), and no part of the curve lies between them.

The empty middle

Put x = 0 to look for points on the y-axis: $0-\frac{y^2}{4}=1$0 − y²/4 = 1, so $y^2=-4$y² = −4. No real number squares to a negative, so the hyperbola never meets the y-axis.

In fact no point of the curve has x between −3 and 3. Rearrange the equation as $\frac{x^2}{9}=1+\frac{y^2}{4}$x²/9 = 1 + y²/4. The term $\frac{y^2}{4}$y²/4 is a square divided by 4, so it is never negative, and $\frac{x^2}{9}$x²/9 is always at least 1. So $x^2$x² is at least 9, which means x is 3 or more, or −3 or less. The whole band between x = −3 and x = 3 is empty, and that gap is what splits the curve into two branches.

Following a branch outward

Walk out along the right-hand branch. At x = 5: $\frac{25}{9}-\frac{y^2}{4}=1$25/9 − y²/4 = 1, so $\frac{y^2}{4}=\frac{25}{9}-1=\frac{16}{9}$y²/4 = 25/9 − 1 = 16/9. Multiply by 4: $y^2=\frac{64}{9}$y² = 64/9, so $y=\frac{8}{3}$y = 8/3 or $y=-\frac{8}{3}$y = −8/3, about 2.67 above and below the axis. At x = 15: $\frac{225}{9}=25$225/9 = 25, so $\frac{y^2}{4}=25-1=24$y²/4 = 25 − 1 = 24, $y^2=96$y² = 96 and y is about 9.80 above and below.

Compare those heights with the straight line $y=\frac{2x}{3}$y = 2x/3. At x = 5 the line is at $\frac{10}{3}$10/3, about 3.33, which is $\frac{10}{3}-\frac{8}{3}=\frac{2}{3}$10/3 − 8/3 = 2/3 above the curve. At x = 15 the line is at 10, only about 0.20 above the curve. The farther out the branch goes, the closer it comes to the line, but it never reaches it. A straight line that a curve approaches more and more closely without meeting is called an asymptote.

The dashed lines are the asymptotes, $y=\frac{2x}{3}$y = 2x/3 and $y=-\frac{2x}{3}$y = −2x/3. Each branch runs closer and closer to them as it moves away from the center.

Finding the asymptotes

Rearrange the equation to find $y^2$y². From $\frac{x^2}{9}-\frac{y^2}{4}=1$x²/9 − y²/4 = 1, $\frac{y^2}{4}=\frac{x^2}{9}-1$y²/4 = x²/9 − 1, and multiplying by 4 gives $y^2=\frac{4x^2}{9}-4$y² = 4x²/9 − 4. When x is large, $\frac{4x^2}{9}$4x²/9 is very large and the − 4 makes almost no difference, so $y^2$y² is very nearly $\frac{4x^2}{9}$4x²/9.

That is the same as replacing the 1 by 0 in the equation: $\frac{x^2}{9}-\frac{y^2}{4}=0$x²/9 − y²/4 = 0. Then $\frac{y^2}{4}=\frac{x^2}{9}$y²/4 = x²/9, so $y^2=\frac{4x^2}{9}$y² = 4x²/9, and taking the square root of both sides gives $y=\frac{2x}{3}$y = 2x/3 or $y=-\frac{2x}{3}$y = −2x/3. Those are the two asymptotes.

In general, the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x²/a² − y²/b² = 1 has its vertices at (a, 0) and (−a, 0), and its asymptotes are $y=\frac{bx}{a}$y = bx/a and $y=-\frac{bx}{a}$y = −bx/a. Take the square roots of the denominators first: here 3 and 2, from 9 and 4, so the gradients are $\frac{2}{3}$2/3 and $-\frac{2}{3}$−2/3, not $\frac{4}{9}$4/9 and $-\frac{4}{9}$−4/9.

Which way it opens

The positive term decides which axis the hyperbola crosses. In $\frac{x^2}{9}-\frac{y^2}{4}=1$x²/9 − y²/4 = 1 the $x^2$x² term is positive, so the curve crosses the x-axis and its branches open to the left and to the right. In $\frac{y^2}{4}-\frac{x^2}{9}=1$y²/4 − x²/9 = 1 the $y^2$y² term is positive: x = 0 gives $y^2=4$y² = 4, so this hyperbola crosses the y-axis at (0, 2) and (0, −2) and its branches open upward and downward. Replacing the 1 by 0 gives the same asymptotes, $y=\frac{2x}{3}$y = 2x/3 and $y=-\frac{2x}{3}$y = −2x/3.

A sum and a difference

Another way to see that the ellipse and the hyperbola are a pair uses two fixed points on the long axis of each curve, called its foci (one focus, two foci). For every point P on an ellipse, the distances from P to the two foci add up to the same total. For every point P on a hyperbola, the two distances differ by the same amount.

The drawing opens on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$x²/25 + y²/16 = 1, whose foci $F_1$F₁ and $F_2$F₂ are at (−3, 0) and (3, 0). Drag P around it: $d_1+d_2$d₁ + d₂ stays 10. Switch to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$x²/9 − y²/16 = 1, whose foci are at (−5, 0) and (5, 0), and the difference of the distances stays 6. At the vertex (3, 0) the distances are 8 and 2, and 8 − 2 = 6.

The usual mistakes

Taking the denominator as the distance. $\frac{x^2}{9}-\frac{y^2}{4}=1$x²/9 − y²/4 = 1 meets the x-axis at x = 3 and x = −3, the square roots of 9, not at $\pm 9$±9.

Turning the gradient upside down. Rooting $y^2=\frac{4x^2}{9}$y² = 4x²/9 gives $y=\pm \frac{2x}{3}$y = ±2x/3: the root of the number under $y^2$y² goes on top. $y=\pm \frac{3x}{2}$y = ±3x/2 is too steep.