Question 768228
Find the minor axis vertices on the ellipse x^2 + 4y^2 + 12x - 32 y +96 = 0?
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x^2+ 12x + 4y^2  - 32 y +96 = 0
complete the square:
(x^2+12x+36)+4(y^2-8y+16)=-96+36+64
(x+6)^2+4(y-4)^2=4
(x+6)^2/4+(y-4)^2=1
This is an equation of an ellipse with horizontal major axis.
Its standard form: {{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}, a>b, (h,k)=(x,y) coordinates of center
For given ellipse:
center: (-6,4)
b^2=1
b=1
minor axis vertices: (-6,4±b)=(-6, 4±1)=(-6,3) and (-6,5)