Given that the ellipse x^2/a^2 + y^2/b^2 =1 has foci at(+/-ae,0)where e is the eccentricity and b^2=a^2(1-e^2),find the coordinates of the foci and the value of e^2 for the ellipse X^2+4y^2 =16
Well you subtract the 16 to get your equation equal to one.
Do x^2/16 + y^2/4=1
Your center is 0,0
Your eclipse would be horizontal.
Do when graphing put a point at (0,0) then go 4 spaces in the negative and positive direction on the x-axis. Then go up 2 spaces from (0,0) in the positive and negative directions on the y-axis. Put your points on the graph. Draw your ellipse.
Then you know to find your c, you do c^2=a^2 -b^2. So that's c^2= 16-4. Which makes c=2radical3
Eccentricity= c/a
So 2radical3/ 4.......e= radical 3 / 2
Foci= (+/- 2radical3, 0)