Question 170163
<pre><font size = 3 color = "indigo"><b>
Yes we can graph on here!

{{{36x^2-288x+25y^2-150y=99}}}

We recognize this as an ellipse because the coefficients of
{{{x^2}}} and {{{y^2}}} when on the same side of the
equation are both positive and unequal. Also the coefficient 
of {{{y^2}}} is smaller than the coefficient of {{{x^2}}}.  
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:

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

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

{{{36(x^2-8x)+25(y^2-6y)=99}}}

Multiply the coefficient of x, which is -8,
by {{{1/2}}}, getting {{{-4}}}.  Then square
{{{-4}}} and get {{{""+16}}}.  Add {{{""+16-16}}},
which amounts to adding zero, inside the first
parentheses:

{{{36(x^2-8x+16-16)+25(y^2-6y)=99}}}

Multiply the coefficient of y, which is -6,
by {{{1/2}}}, getting {{{-3}}}.  Then square
{{{-3}}} and get {{{""+9}}}.  Add {{{""+9-9}}},
which amounts to adding zero, inside the second
parentheses: 

{{{36(x^2-8x+16-16)+25(y^2-6y+9-9)=99}}}

Group the first three terms in each parentheses
and factor:

{{{36((x^2-8x+16)-16)+25((y^2-6y+9)-9)=99}}}

{{{36((x-4)(x-4)-16)+25((y-3)(y-3)-9)=99}}}

Write the factorizations as squares of binomials:

{{{36((x-4)^2-16)+25((y-3)^2-9)=99}}}

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

{{{36(x-4)^2-576+25(y-3)^2-225=99}}}

{{{36(x-4)^2-576+25(y-3)^2-225=99}}}

{{{36(x-4)^2-801+25(y-3)^2=99}}}

{{{36(x-4)^2-801+25(y-3)^2=99}}}

{{{36(x-4)^2+25(y-3)^2=900}}}

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

{{{36(x-4)^2/900+25(y-3)^2/900=900/900}}}

{{{(x-4)^2/25+(y-3)^2/36=1}}}

Comparing this to the form

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

We see that 
{{{matrix(2,4,
h=4, k=3, b^2=25, a^2=36,
 "", "",   b=5,   a=6 )}}}

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

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

{{{drawing(350,400,-3,12,-4.5,10.5, graph(350,400,-3,12,-4.5,10.5),
locate(3.85,3.33,o),locate(4,3,"(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.

{{{drawing(350,400,-3,12,-4.5,10.5, graph(350,400,-3,12,-4.5,10.5),
line(4,3,4,9), locate(3.85,3.33,o),locate(4,3,"(4,3)"),locate(4,9,"(4,9)") )}}}
 
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.

{{{drawing(350,400,-3,12,-4.5,10.5, graph(350,400,-3,12,-4.5,10.5),
line(4,-3,4,9), locate(3.85,3.33,o),locate(4,3,"(4,3)"),locate(4,9,"(4,9)"),
locate(4,-3,"(4,-3)") )}}}

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.

{{{drawing(350,400,-3,12,-4.5,10.5, graph(350,400,-3,12,-4.5,10.5),
line(4,-3,4,9), locate(3.85,3.33,o),locate(4,3,"(4,3)"),locate(4,9,"(4,9)"),
locate(4,-3,"(4,-3)"), line(4,3,9,3),locate(9,3,"(9,3)") )}}}

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:

{{{drawing(350,400,-3,12,-4.5,10.5, graph(350,400,-3,12,-4.5,10.5),
line(4,-3,4,9), locate(3.85,3.33,o),locate(4,3,"(4,3)"),locate(4,9,"(4,9)"),
locate(4,-3,"(4,-3)"), line(-1,3,9,3),locate(9,3,"(9,3)"), locate(-1,3,"(-1,3)") )}}}

Now we can sketch in the graph, approximately:

{{{drawing(350,400,-3,12,-4.5,10.5, graph(350,400,-3,12,-4.5,10.5),
line(4,-3,4,9), locate(3.85,3.33,o),locate(4,3,"(4,3)"),locate(4,9,"(4,9)"),
locate(4,-3,"(4,-3)"), line(-1,3,9,3),locate(9,3,"(9,3)"), locate(-1,3,"(-1,3)"),

graph(350,400,-3,12,-4.5,10.5,3+sqrt(900-36(x-4)^2)/5),
graph(350,400,-3,12,-4.5,10.5,3-sqrt(900-36(x-4)^2)/5)
)}}}
 
)}}}

We need to find the foci:

These are located {{{c}}} units from the center on the major 
axis, where {{{c^2=a^2-b^2}}}.

So we calculate {{{c}}}:

{{{c^2=a^2-b^2}}}
{{{c^2=6^2-5^2}}}
{{{c^2=36-25}}}
{{{c^2=11}}}
{{{c=sqrt(11)}}}

So the foci are the points:

{{{matrix(1,11,
           "(", 4, ",", 3-sqrt(11), ")", and,
           "(", 4, ",", 3+sqrt(11), ")"  )}}} 

The eccentricity is given by the formula {{{e=c/a}}}
{{{e=c/a}}}
{{{e=sqrt(11)/4}}}

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
{{{ matrix(1,3,  y=k+a^2/c, and, y=k-a^2/c)}}}
{{{matrix(1,3,  y=3+36/sqrt(11), and, y=3-36/sqrt(11))}}}
I won't draw these, since they are horizontal lines
both off the graph shown.

Center: {{{matrix(1,5, "(", 4, ",", 3, ")")}}}  
Covertices: {{{matrix(1,11,
           "(", -1, ",", 3, ")", and,
           "(", 9, ",", 3, ")"  )}}} 
Focus or Foci: {{{matrix(1,11,
           "(", 4, ",", 3-sqrt(11), ")", and,
           "(", 4, ",", 3+sqrt(11), ")"  )}}} 
Asymptopes: {{{matrix(1,4, ellipses, have, no, asymptotes)}}}  N/A
Major Axis: {{{matrix(1,9, distance, between, vertices, "=", 2a, "=", 2(6), "=", 12)}}}
Transverse Axis: {{{matrix(1,5, ellipses, have, no, transverse, axis)}}} N/A 
Directrix: If you've studied directrices of ellipse, they are
           {{{matrix(1,3,  y=3+36/sqrt(11), and, y=3-36/sqrt(11))}}}
Vertices/Vertex: {{{matrix(1,11,
           "(", 4, ",", -3, ")", and,
           "(", 4, ",", 9, ")"  )}}} 
Radius: {{{matrix(1,4, ellipses, have, no, radius)}}} N/A
Eccentricity: {{{e=sqrt(11)/4}}}
Minor Axis: {{{matrix(1,9, distance, between, covertices, "=", 2b, "=", 2(5), "=", 10)}}}  
Conjugate Axis: {{{matrix(1,5, ellipses, have, no, 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</pre>