How do you find the standard form of the equation of the ellipse given the properties center (5,2), vertex (0,2) and eccentricity 1/2?

1 Answer
Jan 20, 2018

Answer:

#(x-5)^2/25+(4(y-2)^2)/(75)=1#

Explanation:

Since the center is #(5,2)# and the vertex is #(0,2)# we know two things:

#rarr# #a=5#, because it's the distance from the center to the vertex.
#rarr# the ellipse has a horizontal major axis because the center and vertex are on the same horizontal line.

We can calculate #c#, the distance from the center to the focus, because we know #a# and the eccentricity, #e#, is #e=c/a#:

#1/2=c/5 rarr c = 5/2#.

In an ellipse we know that #c^2=a^2-b^2# and #a=5# while #c=5/2#, so we can find #b#:

#b^2=5^2-(5/2)^2 rarr b^2=25-25/4 rarr b^2=75/4#.

Since the ellipse has a horizontal major axis the standard form of the equation looks like:

#(x-h)^2/a^2+(y-k)^2/b^2=1#

So our equation is:

#(x-5)^2/25+(y-2)^2/(75/4)=1#

or

#(x-5)^2/25+(4(y-2)^2)/(75)=1#