Question 299505
find the vertices, center, and foci of the ellipse, and sketc its graph 9x^2+4y^2-36x+8y+31=0
<pre><font size = 4 color = "indigo"><b>

Rearrange

{{{9x^2-36x+4y^2+8y=-31}}}

Factor out coefficients of squared letters:

{{{9(x^2-4x)+4(y^2+2y)=-31}}}

Complete the square in the first parentheses by
adding {{{red("+4")}}} 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:

{{{9(x^2-4x+red(4))+4(y^2+2y)=-31+red(36)}}}

Complete the square in the second parentheses by
adding {{{red("+1")}}} 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:

{{{9(x^2-4x+red(4))+4(y^2+2y+red(1))=-31+red(36)+red(4)}}}

Factoring the parentheses as a perfect squares:

{{{9(x-2)^2+4(y+1)=9}}}

Get a 1 on the right by dividing through by 9

{{{(9(x-2)^2)/9+(4(y+1)^2)/9=9/9}}}

{{{(x-2)^2/1+4(y+1)^2/9=1}}}

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

{{{(x-2)^2/1+(4(y+1)^2/4)/(9/4)=1}}}

{{{(x-2)^2/1+(cross(4)(y+1)^2/cross(4))/(9/4)=1}}}

{{{(x-2)^2/1+(y+1)^2/(9/4)=1}}}

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

{{{(x-h)^2/b^2+(y-k)^2/a^2=1}}}

{{{h=2}}}, {{{k=-1}}}, 

{{{b^2=1}}} so {{{b=sqrt(1)=1}}}

{{{a^2=9/4}}} so {{{a=sqrt(9/4)=3/2}}}

Its center is at (h,k) = (1,{{{3/2)))))

Plot the center:

{{{drawing(400,400,-2,6,-5,3,
graph(400,400,-2,6,-5,3),
line(2+.1,-1,2-.1,-1), line(2,-1+.1,2,-1-.1),
line(2+.1,-1+.1,2-.1,-1-.1), line(2+.1,-1-.1,2-.1,-1+.1)

  )}}}
 
Draw the major axis {{{a=3/2}}} units both above and below the center.
Draw the minor axis {{{b=1}}} units both right and left of the center.
 

{{{drawing(400,400,-2,6,-5,3,
graph(400,400,-2,6,-5,3), line(2,-2.5,2,1/2), line(1,-1,3,-1) )}}}

The vertices are {{{a=3/2}}} units above and below the center (2,-1)

So we add {{{3/2}}} to the y-coordinate of the center

{{{-1+3/2=-2/2+3/2=1/2}}} so the upper vertex is (2,{{{1/2}}})

And we subtract {{{3/2}}} from the y-coordinate of the center

{{{-1-3/2=-2/2-3/2=-5/2}}} so the lower vertex is (2,{{{-5/2}}})

Sketch in the ellipse:
 
{{{drawing(400,400,-2,6,-5,3,
graph(400,400,-2,6,-5,3), line(2,-2.5,2,1/2), line(1,-1,3,-1), 
arc(2,-1,2,3)
)}}}

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

{{{c^2=a^2-b^2}}}

{{{c^2=(3/2)^2-1^2}}}

{{{c^2=9/4-1}}}

{{{c^2=9/4-4/4}}}

{{{c^2=5/4}}}

{{{c=sqrt(5)/2}}}

The foci are {{{c=sqrt(5)/2}}} units above and below the center (2,-1)

So we add {{{c=sqrt(5)/2}}} to the y-coordinate of the center

{{{-1+sqrt(5)/2}}} so the upper vertex is (2,{{{-1+sqrt(5)/2}}})

And we subtract {{{c=sqrt(13)/2}}} from the y-coordinate of the center

{{{-1-sqrt(5)/2}}} so the upper vertex is (2,{{{-1-sqrt(5/2)}}})

They are approximately 

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

We draw them in



{{{drawing(400,400,-2,6,-5,3,
graph(400,400,-2,6,-5,3), line(2,-2.5,2,1/2), line(1,-1,3,-1), 
arc(2,-1,2,3),

line(2+.1,-1-sqrt(5)/2,2-.1,-1-sqrt(5)/2), line(2,-1-sqrt(5)/2+.1,2,-1-sqrt(5)/2-.1),
line(2+.1,-1-sqrt(5)/2+.1,2-.1,-1-sqrt(5)/2-.1), line(2+.1,-1-sqrt(5)/2-.1,2-.1,-1-sqrt(5)/2+.1),

line(2+.1,-1+sqrt(5)/2,2-.1,-1+sqrt(5)/2), line(2,-1+sqrt(5)/2+.1,2,-1+sqrt(5)/2-.1),
line(2+.1,-1+sqrt(5)/2+.1,2-.1,-1+sqrt(5)/2-.1), line(2+.1,-1+sqrt(5)/2-.1,2-.1,-1+sqrt(5)/2+.1)



)}}}


 
Edwin</pre>