Hyperbola

What is a Hyperbola?

Have you ever seen the shape of cooling towers at power plants? Or perhaps noticed the shadow of a flashlight on the wall forming an open curve? Well, shapes like these are examples of hyperbolas in real life!

A hyperbola is a curve formed when a plane cuts through a double cone at a certain angle. Unlike an ellipse which forms a closed curve, a hyperbola is actually formed from two separate curves that face each other.

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: $PF_1 + PF_2 = 2a$.

Look at the visualization above! A hyperbola has two separate branches. For every point $P$ on the hyperbola, the difference in distance from $P$ to both foci $F_1$ and $F_2$ is always constant.

Components of a Hyperbola

Before diving into formulas, let's get acquainted with the important parts of a hyperbola. Each component has its own role in determining the shape and properties of the hyperbola.

The components of a hyperbola you need to know:

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

2. Foci ($F_1$ and $F_2$) are two fixed points that serve as the reference for the definition of a hyperbola. The distance between the two foci is called the focal distance.

3. Vertices ($A_1$ and $A_2$) are the closest points between the two branches of the hyperbola. The line connecting the two vertices is called the major axis.

4. Major axis is the line that passes through the center and both foci of the hyperbola.

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

There are several forms depending on orientation and center position. We will discuss two cases:

Center at the Origin

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

When the major axis is parallel to the $X$ axis (horizontal), the equation of the hyperbola is:

If we visualize it, it will look like this:

When the major axis is parallel to the $Y$ axis (vertical), the equation of the hyperbola is:

Shifted Center

If the center of the hyperbola is not at the origin, but at point $(h, k)$, the equation becomes:

Important Properties of Hyperbola

There are several mathematical relationships that always hold for every hyperbola:

where:

• $a$ is the distance from center to vertex (semi-major axis)
• $b$ is the constant that determines the shape of the hyperbola (semi-conjugate axis)
• $c$ 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 value of a hyperbola is always $e > 1$. This is because $c > a$ (from the relationship $c^2 = a^2 + b^2$, so $c = \sqrt{a^2 + b^2} > a$). The larger the value of $e$, the more "open" or wide the hyperbola becomes. 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 $(0,0)$:

Asymptotes are "guides" for the hyperbola branches. The farther from the center, the closer the hyperbola curve approaches the asymptote lines, but never touches them.

Exercises

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 major axis, $a = 3$, and $b = 4$. Determine the equation of the hyperbola.

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

Answer Key

1. Solution:

   Given:

   • Center at $(0,0)$
   • Vertices at $(\pm 2, 0)$, so $a = 2$
   • Foci at $(\pm 3, 0)$, so $c = 3$
   • Horizontal major 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 hyperbola with horizontal major 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's in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the major 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 major axis):

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

3. Solution:

   Given:

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

   Equation of hyperbola with center $(h,k)$ and horizontal major 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 major axis (since it's in the form $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$)

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

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

   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:

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

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