Lesson 11.3: Ellipses

New Concept

An ellipse is a conic section defined by two fixed points (foci). This lesson shows you how to graph ellipses, find their equations like $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, and apply these skills to solve real-world problems.

What’s next

Next, you'll dive into interactive examples to master graphing. Then, you'll use these skills to find ellipse equations and solve challenge problems.

Property

An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse. A line drawn through the foci intersects the ellipse in two points, called a vertex of the ellipse. The segment connecting the vertices is called the major axis. The midpoint of the segment is called the center of the ellipse. A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.

Examples

• An ellipse has foci at $(-4, 0)$ and $(4, 0)$. For any point $(x, y)$ on the ellipse, the sum of the distances from that point to each focus is a constant value.
• If an ellipse has vertices at $(6, 0)$ and $(-6, 0)$, its major axis has a length of 12 and its center is at the origin $(0, 0)$.
• If the major axis connects $(0, 7)$ and $(0, -7)$ and the minor axis connects $(-3, 0)$ and $(3, 0)$, the center of the ellipse is their intersection point, $(0, 0)$.

Explanation

Think of an ellipse as a stretched circle with two center points, called foci. The total distance from any point on the curve to both foci is always the same. This constant sum is what creates the unique oval shape.

Property

The standard form of the equation of an ellipse with center $(0, 0)$ is

The $x$-intercepts are $(a, 0)$ and $(-a, 0)$.
The $y$-intercepts are $(0, b)$ and $(0, -b)$.
When $a > b$, the major axis is horizontal. When $b > a$, the major axis is vertical.

Property

1. Write the equation in standard form.
2. Determine whether the major axis is horizontal or vertical.
3. Find the endpoints of the major axis.
4. Find the endpoints of the minor axis.
5. Sketch the ellipse.

Examples

• To graph $\frac{x^2}{16} + \frac{y^2}{4} = 1$, note the major axis is horizontal ($a=4$). The vertices are $(\pm 4, 0)$ and the minor axis endpoints are $(0, \pm 2)$. Sketch the curve through these four points.
• Graph $36x^2 + 4y^2 = 144$. Divide by 144 to get $\frac{x^2}{4} + \frac{y^2}{36} = 1$. The major axis is vertical ($b=6$). The vertices are $(0, \pm 6)$ and minor axis endpoints are $(\pm 2, 0)$.
• An ellipse is described by $\frac{x^2}{49} + \frac{y^2}{81} = 1$. The major axis is vertical ($b=9$). The vertices are $(0, \pm 9)$ and the minor axis endpoints are $(\pm 7, 0)$.

Explanation

To graph an ellipse, first get its equation into standard form. Then, find the four endpoints on the horizontal and vertical axes. These points act as your guide to sketch the smooth, oval shape of the ellipse.

Property

The standard form of the equation of an ellipse with center $(h, k)$ is

When $a > b$, the major axis is horizontal so the distance from the center to the vertex is $a$.
When $b > a$, the major axis is vertical so the distance from the center to the vertex is $b$.

Examples

• The ellipse $\frac{(x-4)^2}{25} + \frac{(y-2)^2}{9} = 1$ has its center at $(4, 2)$. The major axis is horizontal with length $2a=10$. The vertices are at $(4 \pm 5, 2)$, which are $(9, 2)$ and $(-1, 2)$.
• An ellipse is centered at $(-1, 3)$ with a vertical major axis where $b=6$ and a horizontal minor axis where $a=3$. Its equation is $\frac{(x+1)^2}{9} + \frac{(y-3)^2}{36} = 1$.
• For the equation $\frac{(x+5)^2}{100} + \frac{(y+1)^2}{144} = 1$, the center is $(-5, -1)$. Since $b^2 > a^2$, the major axis is vertical.

Explanation

This formula describes an ellipse shifted from the origin. The center is now at $(h, k)$. The values $a$ and $b$ still define the ellipse's size and shape, but the entire figure is moved on the coordinate plane.

Property

To write an equation in standard form, complete the squares for both $x$ and $y$. First, group the $x$ terms and $y$ terms. Factor out the coefficients of the squared terms. Complete the square for each variable, adding the required constants to both sides of the equation. Finally, rewrite as binomial squares and divide by the constant on the right to make the equation equal 1.

Examples

• For $9x^2 + 16y^2 - 18x + 64y - 71 = 0$, completing the square yields $9(x-1)^2 + 16(y+2)^2 = 144$. Dividing by 144 gives the standard form $\frac{(x-1)^2}{16} + \frac{(y+2)^2}{9} = 1$.
• Given $x^2 + 4y^2 + 6x - 8y + 9 = 0$, completing the square results in $(x+3)^2 + 4(y-1)^2 = 4$. Dividing by 4 gives $\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1$.
• Rewrite $3x^2 + 2y^2 - 18x + 4y + 23 = 0$. Completing the square leads to $3(x-3)^2 + 2(y+1)^2 = 6$. Dividing by 6, the standard form is $\frac{(x-3)^2}{2} + \frac{(y+1)^2}{3} = 1$.

Explanation

When an ellipse equation is in a general form, completing the square helps to rewrite it into the neat standard form. This process reveals the ellipse's center, orientation, and axis lengths, making it easy to graph and analyze.

Property

The orbits of planets are elliptical. The closest a planet gets to its sun is one end of the major axis, and the furthest is the other. The sun is at one of the foci. To find the equation of the orbit, we use the relationship $b^2 = a^2 - c^2$, where $2a$ is the total length of the major axis (closest distance + furthest distance) and $c$ is the distance from the center to the focus (the sun).

Examples

• A planet's closest distance to its sun is 30 AU and its furthest is 50 AU. The major axis is $2a = 30+50 = 80$, so $a=40$. The focus is at $c = 40-30=10$. Then $b^2 = 40^2 - 10^2 = 1500$. The orbit is $\frac{x^2}{1600} + \frac{y^2}{1500} = 1$.
• A comet is 10 AU from its sun at its closest and 90 AU at its furthest. The major axis is $2a = 10+90=100$, so $a=50$. The focus is at $c=50-10=40$. Then $b^2=50^2-40^2=900$. The equation is $\frac{x^2}{2500} + \frac{y^2}{900}=1$.
• A planet's orbit has a closest approach of 20 AU and a furthest distance of 60 AU. The major axis is $2a=20+60=80$, so $a=40$. The focus is at $c=40-20=20$. Then $b^2=40^2-20^2=1200$. The equation is $\frac{x^2}{1600} + \frac{y^2}{1200}=1$.

Explanation

Planetary orbits are ellipses, with the sun at a focus, not the center. By knowing the closest and furthest points in an orbit, you can find the major axis length ($2a$), the focus location ($c$), and the minor axis ($b$).