# Hyperbolas in Algebra

These are my notes on hyperbolas in algebra.

**Equations and Graphs**

The third type of conic section is a hyperbola. A hyperbola is the set of points

in a plane, the differences of whose distances from two fixed points is

constant. Each fixed point is called a focus of the hyperbola.

When looking at a hyperbola a point is shown on a hyperbola with distance from

one focus and the distance from the second focus. Regardless of the location of

the point, d2-d1=2a. The transverse axis is the line segment connecting the

vertices and its length equals 2a.

The two parts of the hyperbola are the left branch and the right branch. It can

also be labeled the upper and lower branch. A line segment connecting the points

is the conjugate axis. The lines are asymptotes for each respective hyperbola.

They can be used as an aid in graphing. The dashed rectangle is sometimes called

the fundamental rectangle.

By the vertical line test, a hyperbola cannot be represented by a function, but

many can be described by the following equations. The constants a,b, and c are

positive.

The hyperbola with center at the origin, horizontal transverse axis, and

equation:

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]

has asymptotes \(y = \pm \frac{b}{a} x\), vertices \((\pm a,0)\), where

\(c^2=a^2+b^2\). The hyperbola with center at its origin, vertical transverse

axis, and equation:

\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]

has asymptotes \(y = \pm \frac{a}{b} x\), vertices \((0, \pm a)\), and foci

\((0, \pm c)\), where \(c^2 = a^2+b^2\).

A hyperbola consists of two solid curves or branches. The asymptotes, foci,

transverse axis, and fundamental rectangle are not part of the hyperbola, but

are aids for sketching its graph.**Example 1**

Sketching the graph of a hyperbola

Sketch a graph of \(\frac{x^2}{4} - \frac{y^2}{9} = 1\)

Solution:

The equation is in standard form with a=2 and b=3. It has a horizontal

transverse axis with vertices \((\pm 2,0)\). The endpoints of the conjugate axis

are \((0, \pm 3)\). To locate the foci find c.

\[c^2=a^2+b^2=4+9=13\]

\[c=\sqrt{13} = 3.61\]

The foci are \((\pm \sqrt{13}, 0)\). The asymptotes are \(y=\pm \frac{b}{a} x\)

or \(y=\pm \frac{3}{2} x\).**Example 2**

Finding the equation of a hyperbola

Find the equation of the hyperbola centered at the origin with a vertical

transverse axis of length 6 and focus (0,5).

Solution:

Since the hyperbola is centered at the origin with a vertical axis, its equation

is \(\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\). The transverse axis has length 6=2a,

so a=3. Since one focus is located at (0,5), c=5. We can find b by using the

following equation:

\[b^2=c^2-a^2\]

\[b=\sqrt{c^2-a^2}\]

\[b=\sqrt{5^2-3^2}=4\]

The equation of this hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}=1\). Its

asymptotes are \(y=\pm \frac{a}{b} x\) or \(y=\pm \frac{3}{4} x\).**Example 3**

Finding the equation of a hyperbola

Find the standard equation of the hyperbola. Identify the vertices, foci, and

asymptotes.

Solution:

Because the hyperbola is centered at (0,0) with a horizontal transverse axis,

its equation has the form:

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

The asymptotes are given by \(y=\pm \frac{b}{a}\). The fundamental rectangle is

determined by the four points \((\pm 4,0)\) and \((0,\pm 2)\), and its diagonals

correspond to the asymptotes. It follows that a=4 and b=2. Thus the standard

equation of the hyperbola is:

\[\frac{x^2}{16}-\frac{y^2}{4}=1\]

The vertices are \((\pm 4,0)\), and the asymptotes are \(y=\pm \frac{1}{2} x\).

To find the coordinates of the foci, find c:

\[c^2=a^2+b^2=4^2+2^2=20\]

\[c=\sqrt{20}=4.47\]

The coordinates of the foci are \((\pm \sqrt{20},0)\)**Reflective Properties**

Hyperbolas have an important reflective property. If a hyperbola is rotated

about the x-axis, a hyperboloid is formed. Any beam of light that is directed

toward focus, will be reflected by the hyperboloid toward the other focus.

Telescopes sometimes make use of both parabolic and hyperbolic mirrors. When

parallel rays of light from distant stars strike the large parabolic mirror,

they are reflected toward its focus. A smaller hyperbolic mirror is placed so

that its focus is also located at the first focus. Light rays striking the

hyperbolic mirror are reflected toward is other focus, through a small hole in

the parabolic mirror. **Translations of hyperbolas**

If the equation of a hyperbola is either \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) or

\(\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\), then the center of the hyperbola is

(0,0). We can use the translations of graphs to find the equation of a hyperbola

centered at (h,k) by replacing x with (x-h) and y with (y-k). The constants a,b,

and c are positive.**Example 4**

Graphing a hyperbola with center (h,k)

Graph the hyperbola whose equation \(\frac{(y+2)^2}{9}-\frac{(x-2)^2}{16}=1\).

Solution:

The hyperbola has a vertical transverse axis and its center is (2,-1). Since

\(a^=9\) and \(b^2=16\), it follows that \(c^2=a^2+b^2=9+16=25\). Thus a=3, b=4,

and c=5. The vertices are located 3 units above and below the center of the

hyperbola and the foci are located 5 units above and below the center of the

hyperbola. The vertices are \((2,-1 \pm 3)\) or (2,1) and (2,-5), and the foci

are \((2,-1 \pm 5)\) or (2,3) and (2,-7). The asymptotes are given by:

\[y=\pm \frac{a}{b} (x-h)+k\]

\[y=\pm \frac{3}{4} (x-2)-2\]**Example 5**

Finding the standard equation of the hyperbola

Write \(9x^2-18x-4y^2-16y=43\) in the standard form for a hyperbola centered at

(h,k).

Solution:

We can write the given equation in standard form by completing the square.

\[9x^2-18x-4y^2-16y=43\]

\[9(x^2-2x+1)-4(y^2+4y+4)=43+9-16\]

\[9(x-1)^2-4(y+2)^2=36\]

\[\frac{(x-1)^2}{4}-\frac{(y+2)^2}{9}=1\]

The center is (1,-2). Because a=2 and the transverse axis is horizontal, the

vertices of the hyperbola are \((h \pm a,k)=(1 \pm 2,-2)\).**Solving systems of nonlinear equations**

If the equation of a hyperbola occurs in a system of equations, then the system

of equations is nonlinear.**Example 6**

Solving a system of nonlinear equations

\[4x^2-9y^2=36\]

\[9x^2+25y^2=262\]

Solution:

To eliminate the x-variable, multiply the first equation by -9, the second

equation by 4, and then add the resulting equations.

\[-36x^2+81y^2=-324\]

\[36x^2+100y^2=1048\]

Add these:

\[181y^2=724\]

Now solve:

\[y^2=\frac{724}{181}=4\]

\[y=\pm 2\]

To determine the corresponding x-values, let \(y=\pm 2\) in the given equation

\(4x^2-9y^2=36\).

\[4x^2-9(\pm 2)^2=36\]

\[4x^2-36=36\]

\[4x^2=72\]

\[x^2=18\]

\[x=\pm \sqrt{18}\]

\[x=\pm 3 \sqrt{2}\]

There are four solutions to this system of nonlinear equations:

\[(\pm 3\sqrt{2}, \pm 2)\]