How do you find the center, vertices, foci and eccentricity of x2+4y2+4x+32y+67=0?

1 Answer
Mar 2, 2016

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

Explanation:

Rearranging to the standard form ( x α )^2 / a^2 + ( y β )^2 / b^2 = 1, read semi-major axis a = 1, semi-minor axis b = 1/2, e = ( 1 b^2/a^2 ) = 3 / 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 (α, β ) and foci are at ( α ±ae, β )