The Equation of an Ellipse

Start from a circle

The circle $x^2+y^2=9$x² + y² = 9 has its center at the origin and a radius of 3. Divide both sides by 9, the radius squared: $\frac{x^2}{9}+\frac{y^2}{9}=1$x²/9 + y²/9 = 1. It is the same circle, written so that the right side is 1 and each term has its own denominator.

In this form, each denominator says how far the curve reaches along one axis. Put y = 0, and $\frac{x^2}{9}=1$x²/9 = 1, so $x^2=9$x² = 9 and x = 3 or x = −3. The number under $x^2$x² is the square of the distance from the center to the curve along the x-axis, and the number under $y^2$y² does the same along the y-axis.

Different denominators

Now give the two terms different denominators: $\frac{x^2}{25}+\frac{y^2}{9}=1$x²/25 + y²/9 = 1. The curve is no longer a circle. It is an ellipse, a closed oval with its center at the origin.

Find where it crosses the axes. On the x-axis y = 0, so $\frac{x^2}{25}=1$x²/25 = 1, $x^2=25$x² = 25 and x = 5 or x = −5. On the y-axis x = 0, so $\frac{y^2}{9}=1$y²/9 = 1, $y^2=9$y² = 9 and y = 3 or y = −3. The ellipse crosses the axes at (5, 0), (−5, 0), (0, 3) and (0, −3): it reaches 5 to each side of the center and 3 above and below it.

The ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$x²/25 + y²/9 = 1 reaches 5 along the x-axis on each side of the center, and 3 along the y-axis.

The general equation

An ellipse with its center at the origin has the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x²/a² + y²/b² = 1. It reaches a along the x-axis and b along the y-axis, so its whole width is 2a and its whole height is 2b. For $\frac{x^2}{25}+\frac{y^2}{9}=1$x²/25 + y²/9 = 1, $a^2=25$a² = 25 and $b^2=9$b² = 9, so a = 5 and b = 3: the ellipse is 10 wide and 6 high.

When a = b, the equation is $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$x²/a² + y²/a² = 1. Multiply both sides by $a^2$a² to get $x^2+y^2=a^2$x² + y² = a², which is the circle of radius a. A circle is the ellipse whose two distances are equal.

A stretched circle

Compare the ellipse with the circle $x^2+y^2=25$x² + y² = 25, which reaches 5 in every direction. The point (3, 4) is on that circle, because $3^2+4^2=9+16=25$3² + 4² = 9 + 16 = 25. Now put x = 3 into the ellipse: $\frac{9}{25}+\frac{y^2}{9}=1$9/25 + y²/9 = 1, so $\frac{y^2}{9}=\frac{16}{25}$y²/9 = 16/25. Multiply both sides by 9: $y^2=\frac{144}{25}$y² = 144/25, so $y=\frac{12}{5}=2.4$y = 12/5 = 2.4. Above x = 3 the circle is at height 4 and the ellipse at height 2.4, which is $\frac{3}{5}$3/5 of 4.

The same is true at every x: each height on the ellipse is $\frac{3}{5}$3/5 of the height on the circle. So the ellipse is the circle of radius 5 squashed toward the x-axis to $\frac{3}{5}$3/5 of its height. Stretching a circle by one amount across and a different amount up always gives an ellipse, and the two denominators record the two stretches.

The dashed circle $x^2+y^2=25$x² + y² = 25 and the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$x²/25 + y²/9 = 1 meet at (5, 0) and (−5, 0). Above x = 3 the circle is at height 4 and the ellipse at 2.4, which is $\frac{3}{5}$3/5 as high.

Wide or tall

Swap the two denominators: $\frac{x^2}{9}+\frac{y^2}{25}=1$x²/9 + y²/25 = 1. Now y = 0 gives x = 3 or x = −3, and x = 0 gives y = 5 or y = −5. It is the same ellipse turned upright, 6 wide and 10 high.

So the larger denominator shows which way the ellipse is longer. When the larger one is under $x^2$x², the ellipse is wider than it is tall. When it is under $y^2$y², the ellipse is taller than it is wide. Of all the lines across the ellipse through its center, the longest is called the major axis and the shortest the minor axis.

$\frac{x^2}{9}+\frac{y^2}{25}=1$x²/9 + y²/25 = 1 reaches 3 along the x-axis and 5 along the y-axis, so it stands upright.

Moving the center

To move the center to (2, −1), subtract its coordinates inside the brackets, as in the equation of a circle: $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{4}=1$(x − 2)²/16 + (y + 1)²/4 = 1. The bracket (y + 1) is y − (−1), so the center's y-coordinate is −1.

The denominators still give the distances, now measured from the center. This ellipse reaches $\sqrt{16}=4$√16 = 4 to each side of (2, −1), from x = −2 to x = 6, and $\sqrt{4}=2$√4 = 2 above and below it, from y = −3 to y = 1.

The ellipse $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{4}=1$(x − 2)²/16 + (y + 1)²/4 = 1, centered at (2, −1). It is 8 wide and 4 high.

The usual mistakes

Taking a denominator as the distance. The number under $x^2$x² is $a^2$a², so $\frac{x^2}{25}+\frac{y^2}{9}=1$x²/25 + y²/9 = 1 reaches 5 along the x-axis, not 25.

Matching a denominator to the wrong axis. The number under $x^2$x² belongs to the x-axis. $\frac{x^2}{9}+\frac{y^2}{25}=1$x²/9 + y²/25 = 1 is taller than it is wide, because its larger denominator is under $y^2$y².

Copying the signs out of the brackets. $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{4}=1$(x − 2)²/16 + (y + 1)²/4 = 1 is centered at (2, −1), not (−2, 1): the equation subtracts the center.