How do I find the center, vertices, foci, and eccentricity of the ellipse? #6x^2 + 2y^2 + 18x - 18y + 30 = 0#. I'm really struggling with completing the square.

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Answer 1 · 2017-12-02T21:28:18

See below.

A non slanted ellipse has the structure

#b^2(x-x_0)^2+a^2(y-y_0)^2= a^2b^2# or expanding

#b^2x^2+a^2y^2-2b^2x x_0-2a^2y y_0+b^2x_0^2+a^2y_0^2-a^2b^2-gamma=0#

We introduced an additional parameter #(gamma)# to avoid void solutions in our quest.

Now comparing

#{(b^2=6),(a^2=2),(-2b^2x_0 = 18),(-2a^2 y_0 = -18),(b^2x_0^2+a^2y_0^2-a^2b^2 -gamma= 30):}#

obtaining

#{(b= sqrt6),(a=sqrt2),(gamma = -12),(x_0 = -3/2),(y_0=9/2):}#

So we conclude that

#6x^2 + 2y^2 + 18x - 18y + 30 = 0# does not represent a pure ellipse.

If #6x^2 + 2y^2 + 18x - 18y + 30 = 0# were an ellipse, then #gamma = 0#