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

1 Answer
Oct 27, 2017

Answer:

#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]}