How do you find the a, b, c for #(x + 3)^2/9 + (y – 9)^2/25 =1#?

1 Answer
Oct 22, 2015

Answer:

Find them by a bit of rewriting of the equation and using the formula that #c=sqrt(\mbox{semimajor axis}^2-\mbox{semiminoraxis}^2}#. See below for the answers.

Explanation:

First, rewrite the equation of this ellipse in the form #((x-h)/a)^2+((y-k)/b)^2=1#:

#((x+3)^2)/9+((y-9)^2)/25=1#

#implies ((x-(-3))/3)^2+((y-9)/5)^2=1#.

Hence #a=3# and #b=5#. Since #5>3#, the semimajor axis is #b# and the semiminor axis is #a#.

Therefore, the distance between the center of the ellipse and the two foci is #c=sqrt{5^2-3^2}=sqrt{25-9}=sqrt{16}=4#.

FYI, the center of the ellipse is the point whose rectangular coordinates are #(x,y)=(-3,9)#. The four vertices are the points whose rectangular coordinates are #(x,y)=(-3,9 pm 5)# and #(x,y)=(-3 pm 3,9)#.