SOLUTION: Concerning Ellipses Supposed to find the center, vertices, and foci of the following ellipse. 25X^2+9Y^2+150X-36Y+260=0 Tried to solve and hit a roadblock. solved to the fol

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Concerning Ellipses Supposed to find the center, vertices, and foci of the following ellipse. 25X^2+9Y^2+150X-36Y+260=0 Tried to solve and hit a roadblock. solved to the fol      Log On


   

Question 1173086: Concerning Ellipses
Supposed to find the center, vertices, and foci of the following ellipse.
25X^2+9Y^2+150X-36Y+260=0 Tried to solve and hit a roadblock.
solved to the following:
25(X+3)^2 + 9(Y-2)^2 =1 So center is (-3,2) which agrees with the answer in the book,but here is where I get stuck. My solution for the vertices and foci do not work. Can you help me with the analysis for the vertices and foci? Thank you in advance.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


25x%5E2%2B9y%5E2%2B150x-36y%2B260+=+0 --> 25%28x%2B3%29%5E2%2B9%28y-2%29%5E2+=+1 --> %28x%2B3%29%5E2%2F%28%281%2F5%29%5E2%29%2B%28y-2%29%5E2%2F%28%281%2F3%29%5E2%29+=+1

In the denominators, 1/3 is greater than 1/5, so the semi-major axis a is 1/3 (in the y direction) and the semi-minor axis b is 1/5 (in the x direction).

The vertices are 1/3 unit above and below the center; the co-vertices are 1/5 unit to the right and left of the center.

The foci are c units above and below the center, where so c = 4/15.