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

1 Answer
Bill K.
Oct 22, 2015

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).

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