How do you find the center, vertices, foci and eccentricity of x^2+4y^2+4x+32y+67=0?

1 Answer
A. S. Adikesavan
Mar 2, 2016

Ellipse. C: ( -2, -4 ); A, A', B and B' : ( -1, -4 ), ( -3, -4 ), ( -2, -3.5 ) and ( -2, -4.5 ); e = sqrt3 /2; S ( -2 + sqrt3 / 2, -4 ) and S' ( -2 - sqrt3 / 2 ); ..

Explanation:

Rearranging to the standard form ( x - alpha )^2 / a^2 + ( y -beta )^2 / b^2 = 1, read semi-major axis a = 1, semi-minor axis b = 1/2, e = sqrt( 1 -b^2/a^2 ) = sqrt3 / 2.
The rearranged form is
( x + 2 )^2 + ( y + 4 )^2/(1/4) = 1
This represents an ellipse with axes parallel to axes of coordinates.
Center is (alpha, beta ) and foci are at ( alpha +-ae, beta )

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