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 = #sqrt#3 /2; S ( #-#2 + #sqrt#3 / 2, #-#4 ) and S' ( #-#2 #-# #sqrt#3 / 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 ) = #sqrt#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 (#alpha#, #beta# ) and foci are at ( #alpha# #+-#ae, #beta# )

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