9.4. Ellipses http://www.ck12.org
The focal radius is 3. This means that the foci are at (3, 0) and (-3, 0).
The eccentricity ise=^35.
Example B
Sketch the following ellipse.
(y− 161 )^2 +(x− 92 )^2 = 1
Solution: Plotting the foci are usually important, but in this case the question simply asks you to sketch the
ellipse. All you need is the center,x-radius andy-radius.
Example C
Put the following conic into graphing form.
25 x^2 − 150 x+ 36 y^2 + 72 y− 639 = 0
Solution:
25 x^2 − 150 x+ 36 y^2 + 72 y− 639 = 0
25 (x^2 − 6 x)+ 36 (y^2 + 2 y) = 639
25 (x^2 − 6 x+ 9 )+ 36 (y^2 + 2 y+ 1 ) = 639 + 225 + 36
25 (x− 3 )^2 + 36 (y+ 1 )^2 = 900
25 (x− 3 )^2
900 +36 (y+ 1 )^2
900 =900
900
(x− 3 )^2
36 +(y+ 1 )^2
25 =^1Concept Problem Revisited
Ellipses are measured using their eccentricity. Here are three ellipses with estimated eccentricity for you to compare.