SOLUTION: find the vertices, center, and foci of the ellipse, and sketc its graph 9x^2+4y^2-36x+8y+31=0

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Question 299505: find the vertices, center, and foci of the ellipse, and sketc its graph 9x^2+4y^2-36x+8y+31=0
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
find the vertices, center, and foci of the ellipse, and sketc its graph 9x^2+4y^2-36x+8y+31=0


Rearrange



Factor out coefficients of squared letters:



Complete the square in the first parentheses by
adding  inside the first parentheses
which actually amounts to adding 36 to the left side 
because there is a 9 in front of the parentheses, so
we must add a 36 to the right side:



Complete the square in the second parentheses by
adding  inside the second parentheses
which actually amounts to adding 4 to the left side 
because there is a 4 in front of the parentheses, so
we must add a 4 to the right side:



Factoring the parentheses as a perfect squares:



Get a 1 on the right by dividing through by 9





To get the 4 off the top of the second fraction we 
divide top and bottom by 4:







Since the largest denominator is under the term in
y, the ellipse has a vertical major axis.  So we
compare it to:



, , 

 so 

 so 

Its center is at (h,k) = (1,
 
Draw the major axis  units both above and below the center.
Draw the minor axis  units both right and left of the center.
 



The vertices are  units above and below the center (2,-1)

So we add  to the y-coordinate of the center

 so the upper vertex is (2,)

And we subtract  from the y-coordinate of the center

 so the lower vertex is (2,)

Sketch in the ellipse:
 


To find the foci, we must calculate c, using the Pythagorean
relationship 













The foci are  units above and below the center (2,-1)

So we add  to the y-coordinate of the center

 so the upper vertex is (2,)

And we subtract  from the y-coordinate of the center

 so the upper vertex is (2,)

They are approximately 

(1,0.12) and (1,-2.12)

We draw them in






 
Edwin

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