Hyperbola

The Branches and Foci of a Hyperbola

The vertical outline of a cooling tower often resembles a hyperbola. A conical shock wave from a supersonic aircraft can also trace a hyperbolic path where it intersects the ground. In both cases, the modeled curve has two branches that bend away from a center and approach their asymptotes.

Geometrically, a hyperbola appears when a plane cuts both halves of a double cone. Unlike an ellipse, which is closed, a hyperbola has two separate branches facing away from one another.

Mathematically, a hyperbola is defined as the locus of points where the absolute difference of distances to two fixed points is always constant. These two fixed points are called the foci of the hyperbola. For every point $P$ on the hyperbola, the difference $|PF_1 - PF_2| = 2a$ (constant), whereas in an ellipse the sum of distances is constant: $PF_1 + PF_2 = 2a$.

The visualization shows both branches, the two foci, and one point $P$ on the curve. The two segments have lengths $r_1=PF_1$ and $r_2=PF_2$. Moving $P$ along either branch changes both lengths, but their absolute difference remains $|r_1-r_2|=2a$.

Components of a Hyperbola

Before using the formulas, identify the parts that control the shape and position of the curve.

Read the graph through these five features:

1. Center of hyperbola is the midpoint between the two foci, usually denoted by $O$.

2. Foci ($F_1$ and $F_2$) are the two fixed points in the definition of a hyperbola. If $c$ is the distance from the center to either focus, the distance between the foci is $2c$.

3. Vertices ($A_1$ and $A_2$) are the points where each branch comes closest to the center. They lie on the transverse axis.

4. Transverse axis is the line segment through the center whose endpoints are the two vertices. The foci lie on the line containing this segment.

5. Asymptotes are lines that are approached by the branches of the hyperbola as they extend to infinity. The hyperbola never touches its asymptotes, but the farther from the center, the closer the curve approaches the asymptote lines.

The main difference between a hyperbola and an ellipse: a hyperbola has asymptotes and consists of two separate branches, while an ellipse is a closed curve without asymptotes.

Equation of a Hyperbola

The standard equation depends on the orientation of the branches and the position of the center. We will use two cases. The positive squared term decides the direction, so a positive $x^2$ term opens left and right while a positive $y^2$ term opens up and down.

Center at the Origin

If the center of the hyperbola is at $O(0,0)$, there are two possible orientations:

When the transverse axis is parallel to the $x$ axis, the equation of the hyperbola is:

The corresponding vertical graph is shown below:

When the transverse axis is parallel to the $y$ axis, the equation of the hyperbola is:

Shifted Center

For a hyperbola centered at $(h, k)$, the equation becomes:

For a horizontal transverse axis:

For a vertical transverse axis:

Properties of Hyperbolas

The three parameters of every hyperbola satisfy:

In this notation:

• $a$ is the distance from center to vertex (transverse semiaxis)
• $b$ is the conjugate semiaxis that determines the asymptote slopes
• $c$ is the distance from center to focus

This relationship is different from an ellipse which uses $c^2 = a^2 - b^2$.

Eccentricity of a hyperbola is defined as:

The eccentricity of a hyperbola is always $e > 1$ because $c^2 = a^2 + b^2$ implies $c > a$. More precisely, $e^2 = 1 + \frac{b^2}{a^2}$, so increasing $e$ increases the ratio $b/a$ that also determines the asymptote slopes. For comparison, a circle has $e = 0$, an ellipse has $0 < e < 1$, a parabola has $e = 1$, and a hyperbola has $e > 1$.

Asymptote equations for a hyperbola with center at $(0,0)$:

For a horizontal transverse axis:

For a vertical transverse axis:

The asymptotes describe the limiting directions of the two branches. As the distance from the center grows, each branch approaches an asymptote without intersecting it.

Exercises

Each problem gives a hyperbola equation and asks for its centre, vertices, asymptotes, or foci, so bring the equation into standard form first.

1. Determine the equation of a hyperbola with center at $(0,0)$, vertices at $(\pm 2, 0)$, and foci at $(\pm 3, 0)$.

2. Given the hyperbola $\frac{x^2}{25} - \frac{y^2}{9} = 1$. Determine the coordinates of the foci, eccentricity, and asymptote equations.

3. A hyperbola has its center at $(1, -2)$, horizontal transverse axis, $a = 3$, and $b = 4$. Determine the equation of the hyperbola.

4. Determine the asymptote equations of the hyperbola $\frac{(y+1)^2}{4} - \frac{(x-2)^2}{9} = 1$.

Worked Solutions

1. Solution:

   Given:

   • Center at $(0,0)$
   • Vertices at $(\pm 2, 0)$, so $a = 2$
   • Foci at $(\pm 3, 0)$, so $c = 3$
   • Horizontal transverse axis (since vertices and foci are on the $x$ axis)

   Use the relationship $c^2 = a^2 + b^2$:

   $$3^2 = 2^2 + b^2$$

   $$9 = 4 + b^2$$

   $$b^2 = 5$$

   Equation of a hyperbola with a horizontal transverse axis:

   $$\frac{x^2}{4} - \frac{y^2}{5} = 1$$

2. Solution:

   From the equation $\frac{x^2}{25} - \frac{y^2}{9} = 1$:

   • $a^2 = 25$, so $a = 5$
   • $b^2 = 9$, so $b = 3$

   Since it has the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the transverse axis is horizontal.

   Calculate $c$:

   $$c^2 = a^2 + b^2 = 25 + 9 = 34$$

   $$c = \sqrt{34} \approx 5.831$$

   Coordinates of foci: $(\pm \sqrt{34}, 0)$

   Eccentricity:

   $$e = \frac{c}{a} = \frac{\sqrt{34}}{5} \approx \frac{5.831}{5} \approx 1.166$$

   Asymptote equations (horizontal transverse axis):

   $$y = \pm \frac{b}{a}x = \pm \frac{3}{5}x$$

3. Solution:

   Given:

   • Center: $(h,k) = (1, -2)$
   • Horizontal transverse axis
   • $a = 3$, so $a^2 = 9$
   • $b = 4$, so $b^2 = 16$

   Equation of a hyperbola with center $(h,k)$ and a horizontal transverse axis:

   $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$

   Substituting values:

   $$\frac{(x-1)^2}{9} - \frac{(y+2)^2}{16} = 1$$

4. Solution:

   From the equation $\frac{(y+1)^2}{4} - \frac{(x-2)^2}{9} = 1$:

   • Center: $(h,k) = (2, -1)$
   • $a^2 = 4$, so $a = 2$
   • $b^2 = 9$, so $b = 3$
   • Vertical transverse axis (since it has the form $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$)

   For a hyperbola with center $(h,k)$ and a vertical transverse axis, the asymptote equations are:

   $$y - k = \pm \frac{a}{b}(x - h)$$

   Substituting values:

   $$y - (-1) = \pm \frac{2}{3}(x - 2)$$

   $$y + 1 = \pm \frac{2}{3}(x - 2)$$

   $$y = -1 \pm \frac{2}{3}(x - 2)$$

   So the asymptote equations are:

   $$y = -1 + \frac{2}{3}(x - 2)$$

   $$y = -1 - \frac{2}{3}(x - 2)$$