# How do you write an equation of a ellipse with center (0,4) and a=2c, vertices (-4,4), (4,4)?

Oct 27, 2017

${x}^{2} / 16 + {\left(y - 4\right)}^{2} / 12 = 1$

#### Explanation:

The general form for an ellipse is either of the following:

$\left.\begin{matrix}{\left(x - {x}_{c}\right)}^{2} / {a}^{2} + {\left(y - {y}_{c}\right)}^{2} / {b}^{2} = 1 & \text{Horizontal" \\ (y-y_c)^2/a^2 + (x-x_c)^2/b^2 = 1 & "Vertical}\end{matrix}\right.$

$\text{Center: } \left({x}_{c} , {y}_{c}\right)$

In either case, the quantity ${a}^{2}$ is strictly greater than ${b}^{2}$.

By convention of writing for all ellipses, the value $a$ represents the semi-major axis length, and the value $b$ represents the semi-minor axis length. Furthermore, there is a relationship between both of those values that satisfies this relationship:

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

Although we do not know whether this is a horizontal or vertical ellipse, we can use this relationship to determine another fact:

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

${c}^{2} = {\left(2 c\right)}^{2} - {b}^{2}$

${c}^{2} = 4 {c}^{2} - {b}^{2}$

${b}^{2} = 3 {c}^{2}$

We know from looking at the vertices of (-4, 4) and (4,4) provided, and the center of (0, 4), that the horizontal semi-axis length is 4. This is either the semi-major axis (if the ellipse is horizontally aligned), or the semi-minor axis.

At this point, we can choose which kind of ellipse we want to find. I will assume this is a horizontal ellipse and omit a solution for a vertical ellipse. If this is a horizontal ellipse, than the $a$ value will correspond to the provided horizontal semi-axis length of 4. Thus:

$a = 4$

$a = 2 c \implies c = \frac{a}{2} \therefore c = 2$

${b}^{2} = 3 {c}^{2} \implies {b}^{2} = 3 \cdot 4 = 12 \therefore b = \sqrt{12} = 2 \sqrt{3}$

Putting all of this together, and using the horizontal ellipse equation, gives us:

${\left(x - 0\right)}^{2} / {\left(4\right)}^{2} + {\left(y - 4\right)}^{2} / {\left(2 \sqrt{3}\right)}^{2} = 1$

${x}^{2} / 16 + {\left(y - 4\right)}^{2} / 12 = 1$

Graph:

graph{x^2/16 + (y-4)^2/12 = 1 [-6, 6, -2, 8]}