This is a conics question for my college algebra class.
How do I find the center, vertices, and foci of the following equation for an ellipse? X^2 + 3y^2 - 12y +9 = 0
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X^2 + 3y^2 - 12y +9 = 0; The general form of the equation of an ellipse is Ax^2 + Cy^2 + Dx + Ey + F =0
Then in the equation that we have A= 1; C= 3; D =0; E=-12 and F=9. To solve the parts of the ellipse we arrange: a=(distance from vertice to center) = √C =√3; half distance of minor axis =b=√A =√1=1, the center is on point (h,k) h is defined as (-D/2A)=0 and k is (-E/2C) = -(-12/6)=2 the center is at the point (0,2). The foci will be c from the center and c = √(a^2 - b^2) =√(3-1) = √2. the distance to the vertice is a=√(3). the points asked are:
Foci ( -√2, 2) and (√2,2)
Vertices (√3,2) and (-√3,3)