SOLUTION: List all Applicable Information for the conic given in standard form. If the information isn't needed for the conic, write N/A. Show Work. (We have to graph it, but i dont think we

Yes we can graph on here!



We recognize this as an ellipse because the coefficients of
 and  when on the same side of the
equation are both positive and unequal. Also the coefficient 
of  is smaller than the coefficient of .  
So the graph will looks like the number 0 (zero).  That is, 
its major axis will be vertical.

We must get it in the standard form:



Factor 36 out of the first two terms on the left,
and factor 25 out of the last two terms on the
left:



Multiply the coefficient of x, which is -8,
by , getting .  Then square
 and get .  Add ,
which amounts to adding zero, inside the first
parentheses:



Multiply the coefficient of y, which is -6,
by , getting .  Then square
 and get .  Add ,
which amounts to adding zero, inside the second
parentheses: 



Group the first three terms in each parentheses
and factor:





Write the factorizations as squares of binomials:



Distribute the 36 and the 25 into the outer
sets of parentheses, leaving the squares of
binomials intact:











Get a 1 on the right by dividing every term
by 900:





Comparing this to the form



We see that 


So the center is the point (h,k) = (4,3)

Let's plot the center (4,3):

 

Now since the ellipse's major axis is vertical, and since a = 6,
draw a vertical line from the center extending 6 units upward,
which will put it end at the point (4,9).  That will be one of 
the vertices.


 
Also draw another vertical line from the center extending 6 units
downward, which will put its end at the point (4,-3) which will be
the other vertex.



since b = 5, draw a horizontal line from the center extending 5 units to the
right, which will put its end at the point (9,3).  That will be one of 
the covertices.



Also draw a horizontal line from the center extending 5 units to the
left, which will put its end at the point (-1,3).  That will be the 
other covertex:



Now we can sketch in the graph, approximately:


 
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We need to find the foci:

These are located  units from the center on the major 
axis, where .

So we calculate :







So the foci are the points:

 

The eccentricity is given by the formula 



You may not have studied that ellipses have 
two directrices. Anyway, for an ellipse elongated
vertically the equations of the two directrices
of the ellipse is given by the equation


I won't draw these, since they are horizontal lines
both off the graph shown.

Center:   
Covertices:  
Focus or Foci:  
Asymptopes:   N/A
Major Axis: 
Transverse Axis:  N/A 
Directrix: If you've studied directrices of ellipse, they are
           
Vertices/Vertex:  
Radius:  N/A
Eccentricity: 
Minor Axis:   
Conjugate Axis:  N/A 
Axis of Symmetry: There are two axes of symmetry. The horizontal axis of
                  symmetry is the line y=3 and the vertical line of 
                  symmetry is the line x=4.

Edwin