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Conics: Standard form equation question.

The directions are the write the standard form equation of the following conic section in which the foci are (6,1), (6,-11) and eccentricity = 3/5.
Please explain how you get the denominator of the answer.  I don't understand how to get the denominator.  Thank you!

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Francisco P. | Summer Physics TutorSummer Physics Tutor
5.0 5.0 (212 lesson ratings) (212)
For a conic, the eccentricity (e) between 0 and 1 will make this an ellipse.
Since the foci are located at (6,1) and (6,-11), the distance between foci is 12 units = 2c.
The eccentricity is related to half of this distance and half of the length of the major axis of the ellipse (2a).
e = c/a
We also have a2-b2 = c2 for an ellipse.
So c = 6, which makes a = 10, which makes b = 8.
Since the center of the ellipse is at (6,-5), the equation for the ellipse is (y+5)2/a2 + (x-6)2/b2 = 1
or (y+5)2/100+(x-6)2/64=1