SOLUTION: Find the minor axis vertices on the ellipse x^2 + 4y^2 + 12x - 32 y +96 = 0? I don't understand how to do this, thanks in advance for your help!

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Question 768228: Find the minor axis vertices on the ellipse x^2 + 4y^2 + 12x - 32 y +96 = 0? I don't understand how to do this, thanks in advance for your help!

Found 2 solutions by lwsshak3, MathLover1:
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
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: , 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)

Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!

.....rearrange terms

.......complete the square, write as
then write as
....group

.......both sides divide by

and ,
and

The distance from the center to either focus is the fixed value . The distance from the center to a vertex is the fixed value .
so center is at (,)
the length of the whole major axis is
the length of the whole minor axis is
vertices: (,) and (,)