Standard Form of the Equation

Key Questions

How do I know whether the major axis of an ellipse is horizontal or vertical?

Locate the larger of the 2 denominators. Then look at that numerator.

If it has an $x$ then that is the major axis. Horizontal
If it has an $y$ then that is the major axis. Vertical

$x^2/4+y^2/16=1$

The major axis is $y$ . Vertical

$x^2/25+y^2/16=1$

The major axis is $x$ . Horizontal

AJ Speller · 11 · Sep 28 2014
What is meant by an ellipse in standard form?

The standard form of the ellipse, centered in the point $C(x_C,y_C)$ and with the semi-axes $a$ , horizontal and $b$ , vertical is:

$(x-x_C)^2/a^2+(y-y_C)^2/b^2=1$ .

Massimiliano · 2 · Mar 2 2015
What do $a$ and $b$ represent in the standard form of the equation for an ellipse?

For ellipses, $a >= b$ (when $a = b$ , we have a circle)

$a$ represents half the length of the major axis while $b$ represents half the length of the minor axis.

This means that the endpoints of the ellipse's major axis are $a$ units (horizontally or vertically) from the center $(h, k)$ while the endpoints of the ellipse's minor axis are $b$ units (vertically or horizontally)) from the center.

The ellipse's foci can also be obtained from $a$ and $b$ .
An ellipse's foci are $f$ units (along the major axis) from the ellipse's center

where $f^2 = a^2 - b^2$

Example 1:

$x^2/9 + y^2/25 = 1$

$a = 5$
$b = 3$

$(h, k) = (0, 0)$

Since $a$ is under $y$ , the major axis is vertical.

So the endpoints of the major axis are $(0, 5)$ and $(0, -5)$

while the endpoints of the minor axis are $(3, 0)$ and $(-3, 0)$

the distance of the ellipse's foci from the center is

$f^2 = a^2 - b^2$

$=> f^2 = 25 - 9$
$=> f^2 = 16$
$=> f = 4$

Therefore, the ellipse's foci are at $(0, 4)$ and $(0, -4)$

Example 2:

$x^2/289 + y^2/225 = 1$

$x^2/17^2 + y^2/15^2 = 1$

$=> a = 17, b = 15$

The center $(h, k)$ is still at (0, 0).
Since $a$ is under $x$ this time, the major axis is horizontal.

The endpoints of the ellipse's major axis are at $(17, 0)$ and $(-17, 0)$ .

The endpoints of the ellipse's minor axis are at $(0, 15)$ and $(0, -15)$

The distance of any focus from the center is

$f^2 = a^2 - b^2$
$=> f^2 = 289 - 225$
$=> f^2 = 64$
$=> f = 8$

Hence, the ellipse's foci are at $(8, 0)$ and $(-8, 0)$

seph · 1 · Oct 16 2014
How do I use completing the square to rewrite the equation of an ellipse in standard form?

To demonstrate, let's try transforming the equation below into standard form.

$4x^2 + 9y^2 + 72y - 24x + 144 = 0$

The first thing to do is group $x$ s and $y$ s together

$(4x^2 - 24x) + (9y^2 + 72y) + 144$

Factor out $x^2$ 's and $y^2$ 's coefficient

$4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 0$

Before we continue, let's recall what happens when a binomial is squared

$(ax + b)^2 = a^2x^2 + 2abx + b^2$

For our problem, we want $x^2 - 6x$ and $y^2 + 8y$ to be perfect squares

For $x$ , we know that

$a^2 = 1$
$=> a = 1, a = -1$

$2ab = -6$
$2(1)b = -6$
$2b = -6$
$b = -3$
$=> b^2 = 9$

Note that substituting $a = -1$ will result to the same $b^2$

Hence, to complete the square, we need to add $9$

Meanwhile, for $y$

$a^2 = 1$
$=> a = 1, a = -1$

$2ab = 8$
$2(1)b = 8$
$2b = 8$
$b = 4$
$=> b = 16$

Now, let's add our $b^2$ s into the equation.

$4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 0$

$4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0$

We added something into the left-hand side, to retain the "equality", we need to add the same value to the right-hand side (or substract the value again from the left-hand side)

$4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0 + 4(9) + 9(16)$
$4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 180$

$=> 4(x - 3)^2 + 9(y + 4)^2 + 144 = 180$
$=> 4(x - 3)^2 + 9(y + 4)^2 = 180 - 144$
$=> 4(x - 3)^2 + 9(y + 4)^2 = 36$

$=> (4(x - 3)^2 + 9(y + 4)^2 = 36)/36$

$=> (x - 3)^2/9 + (y + 4)^2/4 = 1$

seph · · Oct 19 2014

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Geometry of an Ellipse

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Standard Form of the Equation

Key Questions

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Geometry of an Ellipse