How do you find the equation of an ellipse with foci #(+-2,0)# and major axis of length 8?

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Answer 1 · 2016-12-17T02:09:52

#x^2/16+y^2/12=1#. Graph is inserted.

The line of foci is the major axis and, here, it is the x-axis y = 0.

The center is the midpoint of the join of theci, and so, it is the origin

(0, 0 ).

The major axis 2a = 8. So, a = 4.

THe distance between the foci = 2a X (eccentricity) = 2ae = 6e = 4. So, e = 1/2.

The semi minor axis #b = aasqrt(1-e^2)=4sqrt(1-1/4)=2sqrt3#.

Now, the equation is

#x^2/a^2+y^2/b^2=x^2/16+y^2/12=1#

The graph is inserted.

graph{x^2/16+y^2/12-1=0 [-10, 10, -5, 5]}