SOLUTION: Find the center,foci, vertices and lengh\th of the axes for the ellipse Equation:4x^2+y^2+24x-10y+45=0

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Question 856394: Find the center,foci, vertices and lengh\th of the axes for the ellipse
Equation:4x^2+y^2+24x-10y+45=0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
4x%5E2%2By%5E2%2B24x-10y%2B45=0
4x%5E2%2B24x%2By%5E2-10y=-45
4x%5E2%2B24x%2B36%2By%5E2-10y%2B25=-45%2B36%2B25
4%28x%5E2%2B6x%2B9%29%2B%28y-5%29%5E2=16
4%28x%2B3%29%5E2%2B%28y-5%29%5E2=16
4%28x%2B3%29%5E2%2F16%2B%28y-5%29%5E2%2F16=1
%28x%2B3%29%5E2%2F4%2B%28y-5%29%5E2%2F16=1
%28x%2B3%29%5E2%2F2%5E2%2B%28y-5%29%5E2%2F4%5E2=1
That is the equation of an ellipse
centered at highlight%28%22%28+-3+%2C+5+%29%22%29 ,
with major axis parallel to the y-axis
and semi-major axis a=4 ,
with minor axis parallel to the x-axis
and semi-major axis b=2 ,
and a focal distance c such that
b%5E2%2Bc%5E2=a%5E2 so
2%5E2%2Bc%5E2=4%5E2 or 4%2Bc%5E2=16 so
c%5E2=16-4-->c%5E2=12-->c=sqrt%2812%29-->c=2sqrt%283%29--> c=about3.464

The vertices are 4 units above and 4 units below the center,
at highlight%28%22%28+-3+%2C+9+%29%22%29 and highlight%28%22%28+-3+%2C+1+%29%22%29 .
The foci are 2sqrt%283%29 above and 2sqrt%283%29 below the center,
at highlight%28%22%28+-3+%2C+5%2B2+sqrt%283%29%29%22%29 and highlight%28%22%28+-3+%2C+5-2+sqrt%283%29%29%22%29 .
The length of the major axis is the distance between the vertices
2a=2%2A4=highlight%288%29 .
The length of the minor axis is
2b=2%2A2=highlight%284%29 .