# Standard Form of the Equation

## Key Questions

• 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

• The standard form of the ellipse, centered in the point $C \left({x}_{C} , {y}_{C}\right)$ and with the semi-axes $a$, horizontal and $b$, vertical is:

${\left(x - {x}_{C}\right)}^{2} / {a}^{2} + {\left(y - {y}_{C}\right)}^{2} / {b}^{2} = 1$.

• For ellipses, $a \ge b$ (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 $\left(h , k\right)$ 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:

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

$a = 5$
$b = 3$

$\left(h , k\right) = \left(0 , 0\right)$

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

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

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

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

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

$\implies {f}^{2} = 25 - 9$
$\implies {f}^{2} = 16$
$\implies f = 4$

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

Example 2:

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

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

$\implies a = 17 , b = 15$

The center $\left(h , k\right)$ 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 $\left(17 , 0\right)$ and $\left(- 17 , 0\right)$.

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

The distance of any focus from the center is

${f}^{2} = {a}^{2} - {b}^{2}$
$\implies {f}^{2} = 289 - 225$
$\implies {f}^{2} = 64$
$\implies f = 8$

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

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

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

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

$\left(4 {x}^{2} - 24 x\right) + \left(9 {y}^{2} + 72 y\right) + 144$

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

$4 \left({x}^{2} - 6 x\right) + 9 \left({y}^{2} + 8 y\right) + 144 = 0$

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

${\left(a x + b\right)}^{2} = {a}^{2} {x}^{2} + 2 a b x + {b}^{2}$

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

For $x$, we know that

${a}^{2} = 1$
$\implies a = 1 , a = - 1$

$2 a b = - 6$
$2 \left(1\right) b = - 6$
$2 b = - 6$
$b = - 3$
$\implies {b}^{2} = 9$

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

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

Meanwhile, for $y$

${a}^{2} = 1$
$\implies a = 1 , a = - 1$

$2 a b = 8$
$2 \left(1\right) b = 8$
$2 b = 8$
$b = 4$
$\implies b = 16$

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

$4 \left({x}^{2} - 6 x\right) + 9 \left({y}^{2} + 8 y\right) + 144 = 0$

$4 \left({x}^{2} - 6 x + 9\right) + 9 \left({y}^{2} + 8 y + 16\right) + 144 = 0$

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)

$4 \left({x}^{2} - 6 x + 9\right) + 9 \left({y}^{2} + 8 y + 16\right) + 144 = 0 + 4 \left(9\right) + 9 \left(16\right)$
$4 \left({x}^{2} - 6 x + 9\right) + 9 \left({y}^{2} + 8 y + 16\right) + 144 = 180$

$\implies 4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} + 144 = 180$
$\implies 4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} = 180 - 144$
$\implies 4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} = 36$

$\implies \frac{4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} = 36}{36}$

$\implies {\left(x - 3\right)}^{2} / 9 + {\left(y + 4\right)}^{2} / 4 = 1$