# Ellipses in Algebra

These are my notes on ellipses in algebra.

One method for sketching an ellipse is to tie a string to two nails driven into

a flat board. If a pencil is placed inside the loop formed by the string, the

resulting curve is an ellipse. The sum of the distances \(d_{1}\) and \(d_{2}\)

between the pencil and each of the nails is always fixed by the string. The

locations of the nails correspond to the foci of the ellipse. If the two nails

coincide, the ellipse becomes a circle. As the nails spread farther apart, the

ellipse becomes more elongated, or eccentric.

An ellipse is the set of points in a plane, the sum of whose distances from two

fixed points is constant. Each fixed point is called a focus of the ellipse.

The major axis and minor axis are labeled for each ellipse. The major axis is

the line segment connecting V1 and V2, and the minor axis is the line segment

connecting U1 and U2.

Since a vertical line can intersect the graph of an ellipse more than once, an

ellipse cannot be described by a function. **Standard Equations for Ellipses**

The ellipse with center at the origin, horizontal major axis, and equation

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

has vertices \((\pm a,0)\), endpoints of the minor axis \((0, \pm b)\), and foci

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

The ellipse with center at the origin, vertical major axis, and equation

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

has vertices \((0, \pm a)\), endpoints of the minor axis \((\pm b,0)\) and foci

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

If a=b, then the ellipse is a circle with radius r=a and center (0,0).**Example 1**

Sketching graphs of ellipses

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

Solution:

This equation describes an ellipse with a=3 ands b=2. The ellipse has a

horizontal major axis with vertices \((\pm 3,0)\). The endpoints of the minor

axis are \((0,\pm 2)\). to locate the foci, find c:

\[c^2=a^2-b^2=9-4=5=\sqrt{5}=2.24=c\]**Example 2**

Finding the equation of an ellipse

Solution:

The ellipse is centered at (0,0) and has a horizontal major axis. Its standard

equation has the form:

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

The endpoints of the major axis are \((\pm 4,0)\), and the endpoints of the

minor axis are:

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

The foci lie on the horizontal major axis and can be determined as follows:

\[c^2=a^2-b^2=16-4=12\]

So, \(c=\sqrt{12}=3.46\), and the coordinates of the foci are \((\pm

\sqrt{12},0)\). **Example 3**

Finding the equation of an ellipse

Find the standard equation of the ellipse with foci \((0,\pm 1)\) and vertices

\((0,\pm 3)\).

Solution:

Since the foci and the vertices lie on the y-axis, the ellipse has a vertical

major axis. Its standard equation has the form:

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

Because the foci are \((0,\pm 1)\) and the vertices are \((0,\pm 3)\), it

follows that c=1 and a=3. The value of \(b^2\) can be found using the equation

\(c^2=a^2-b^2\).

\[b^2=a^2-c^2=9-1=8\]

The equation of the ellipse is:

\[\frac{x^2}{8}+\frac{y^2}{9}=1\]

To graph the ellipse, it is helpful to note that the endpoints of the minor axis

are \((\pm b,0)\) or \((\pm \sqrt{8},0)\) where \(\sqrt{8}=2.83\).

The planets travel around the sun in elliptical orbits. Although their orbits

are nearly circular, many planets have a slight eccentricity to them. The

eccentricity e of an ellipse is defined by:

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

Since the foci of an ellipse lie inside the ellipse, \(0 \leq c < a\) and \(0

\leq c/a <1\). Therefore the eccentricity e of an ellipse satisfies \(0 \leq e

<1\). If e=0, then a=b and the ellipse is a circle. Astronomers have measured

values of a and e for each planet. With this information and the fact that the

sun is located at one focus of the ellipse, the equation of a planet's orbit can

be found. **Example 4**

Finding the orbital equation of the planet Pluto.

the planet Pluto has a=39.44 and e=0.249, the greatest eccentricity of any

planet. Graph the orbit of Pluto and the position of the sun in [-60,60,10] by

[-40,40,10].

Solution:

Let the orbit of Pluto be given by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

\[e=\frac{c}{a}=0.249\]

This implies that :

\[c=0.249a=0.249(39.44)=9.821\]

To find b, solve the equation \(c^2=a^2-b^2\) for b.

\[b=\sqrt{a^2-c^2}=\sqrt{39.44^2-9.821^2}=38.20\]

Pluto's orbit is modeled by \(x^2/39.44^2=1\). Since c=9.821, the foci are

\((\pm 9.821,0)\). The sun could be located at either focus. We locate the sun

at (9.821,0).**Reflective Properties of Ellipses**

Like parabolas, ellipses also have an important reflective property. If an

ellipse is rotated about the x-axis, an ellipsoid is formed, which resembles the

shell of an egg. If a light source is placed at focus f, then every beam of

light emanating from the light source, regardless of its direction, is reflected

at the surface of the ellipsoid toward focus f2.**Translations of Ellipses**

If the equation of an ellipse is given by either:

\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ or } \frac{x^2}{b^2}+\frac{y^2}{a^2}=1\]

then the center of the ellipse is (0,0). We can use translations of graphs to

find the equation of an ellipse centered at (h,k) by replacing x with (x-h) and

y with (y-k).**Example 5**

Translating an ellipse

Translate the ellipse with equation \(\frac{x^2}{9}+\frac{y^2}{4}=1\) so that it

is centered at (-1,2). Find this equation.

Solution:

To translate the center from (0,0) to (-1,2), replace x with (x+1) and y with

(y-2). This new equation is:

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

This ellipse is congruent to the given ellipse that is centered at (-1,2). **Example 6**

Graphing an ellipse with center (h,k)

Graph the ellipse whose equation is:

\[\frac{(x+2)^2}{16}+\frac{(y-2)^2}{25}=1\]

Solution:

The ellipse has a vertical major axis and its center is (-2,2). Since \(a^2=25\)

and \(b^2=16\), it follows that \(c^2=a^2-b^2=25-16=9\). So, a=5, b=4, and c=3.

The vertices are located 5 units above and below the center of the ellipse and

the foci are located 3 units above and below the center of the ellipse. The

vertices are \((-2,2 \pm 5)\), or (-2,7) and (-2,-3(, and the foci are \((-2,2

\pm 3)\), or (-2,5) and (-2,-1). **Example 7**

Finding the standard equation of an ellipse

Write \(4x^2-16x+9y^2+54y+61=0\) in the standard form for an ellipse centered at

(h,k).

Solution:

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

\[4x^2-16x+9y^2+54y+61=0\]

\[4(x^2-4x+ )+9(y^2+6y+ )=-61\]

\[4(x^2-4X+4)+9(Y^2+6Y+9)=-61+16+81\]

\[4(X-2)^2+9(Y+3)^2=36\]

\[\frac{(x-2)^2}{9}+\frac{(y+3)^2}{4}=1\]

The center is (2,-3). Because a=3 and the major axis is horizontal, the vertices

of the ellipse are \((2 \pm 3,-3)\) or (5,-3) and (-1,-3).**Circles**

A circle is an ellipse where a=b. If a circle has radius r and center (h,k),

then an equation for the circle is:

\[\frac{(x-h)^2}{r^2}+\frac{(y-k)^2}{r^2}=1\]

multiplying the equation by \(r^2\) provides the following:

\[(x-h)^2+(y-k)^2=r^2\]

This is the standard equation of a circle.**Example 8**

Finding the standard equation of a circle

Find the standard equation of a circle with radius 4 and center (5,-3)

Solution:

Let h=5, k=-3, and r=4

The standard equation is:

\[(x-5)^2+(y+3)^2=16\]**Example 9**

Finding the center and radius of a circle

Find the center and radius of the circle given by:

\[x^2+6x+y^2-2y=-6\]

Solution:

Complete the square to write the standard equation of the circle.

\[x^2+6x+y^2-2y=-6\]

\[(x^2+6x+9)+(y^2-2y+1)=-6+9+1\]

\[(x+3)^2+(y-1)^2=4\]

The center is (-3,1) and the radius is 2**Area Inside an Ellipse**

Given the standard equation of an ellipse, the area A of the region contained

inside is given by \(A=\pi ab\)**Example 11**

Finding the area inside an ellipse

Shade the region in the xy-plane that satisfies the inequality \(x^2+4y^2\leq

4\). Find the area of this region if units are in inches.

Solution:

Begin by dividing each term in the given inequality by 4

\[x^2+4y^2 \leq 4\]

\[\frac{x^2}{4}+\frac{4y^2}{4}\leq\frac{4}{4}\]

\[\frac{x^2}{4}+\frac{y^2}{1}\leq1\]

The boundary of the region is the ellipse:

\[\frac{x^2}{4}+\frac{y^2}{1}\leq1\]

The region inside the ellipse satisfies the inequality. To verify, note that the

test point (0,0), which is located inside the ellipse, satisfies the inequality.

The solution is shaded and the area of this region is:

\[A=\pi ab=\pi (2)(1)=2\pi=6.28\]