Question

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

Answer 1

`x^2/16 + (y-4)^2/12 = 1`

Explanation:

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

`{:((x-x_c)^2/a^2 + (y-y_c)^2/b^2 = 1,"Horizontal"),((y-y_c)^2/a^2 + (x-x_c)^2/b^2 = 1, "Vertical") :}`

`"Center: "(x_c,y_c)`

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 = (2c)^2 - b^2`

`c^2 = 4c^2 - b^2`

`b^2 = 3c^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 = 2c => c = a/2 :. c = 2`

`b^2 = 3c^2 => b^2 = 3*4 = 12 :. b = sqrt(12) = 2sqrt(3)`

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

`(x-0)^2/(4)^2 + (y-4)^2/(2sqrt(3))^2 = 1`

`x^2/16 + (y-4)^2/12 = 1`

Graph:

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