The first thing we want to do is put the conic (an ellipse because the x² and the y² terms have the same sign) into a better form i.e. 

\(\displaystyle \frac{\left ( x-h\right )^2}{a^2}+\frac{\left ( y-k\right )^2}{b^2}=1\)

where (h,k) is the center of our ellipse.

We will continue by completing the square for both the x and y binomials.

First we seperate them into two trinomials:

  \(\displaystyle \left ( 27x^2+324x+0\right )+\left ( 16y^2-96y+0\right )=-684\)

then we pull a 27 out of the first one and a 16 out of the second

\(\displaystyle 27\left ( x^2+12x+0\right )+16\left ( y^2-6y+0\right )=-684\)

then we add the correct constant to each trinomial (being sure to add the same amount to the other side of our equation.

\(\displaystyle 27\left ( x^2+12x+36\right )+16\left ( y^2-6y+9\right )=-684+27(36)+16(9)\)

then we factor our trinomials and divide by 16 and 27 to get

\(\displaystyle \frac{\left ( x+6\right )^2}{16}+\frac{\left ( y-3\right )^2}{27}=1\)

so the center of our ellipse is (-6,3) and we calculate the distance from the focal points to the center by the equation:

 \(\displaystyle c^2=\left | a^2-b^2\right |\)

and we know that our ellipse is stretched in the y direction because b>a so our focal points will be c displaced from our center. 

with \(\displaystyle c=\sqrt{11}\)

our focal points are \(\displaystyle \left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )\)

To find the center of this ellipse we need to put it into a better form. We do this by rearranging our terms and completing the square for both our y and x terms.

\(\displaystyle 33(x^2-12x)+13(y^2+2\sqrt7y)=-850\) 

completing the square for both gives us this.

\(\displaystyle 33(x^2-12x+36)+13(y^2+2\sqrt7y+7)=429\)

\(\displaystyle 33(x-6)^2+13(y+\sqrt7)^2=429\)

we could divide by 429 but we have the information we need. The center of our ellipse is \(\displaystyle (6,-\sqrt7 )\)

The equation of an ellipse is

\(\displaystyle \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\),

where a is the horizontal radius, b is the vertical radius, and (h, k) is the center of the ellipse. In this case we are told that the center is at the origin, or (0,0), so both h and k equal 0. That brings us to:

\(\displaystyle \frac{(x-0)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1\)

\(\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

We are told about the major and minor radiuses, but the problem does not specify which one is horizontal and which one vertical. However it does tell us that the ellipse passes through the point (5, 0), which is in a horizontal line with the center, (0, 0). Therefore the horizontal radius is 5.

The vertical radius must then be 3. We can now plug these in:

\(\displaystyle \frac{x^2}{5^2} + \frac{y^2}{3^2} = 1\)

\(\displaystyle \frac{x^2}{25} + \frac{y^2}{9} = 1\)

The usual form for an ellipse is

\(\displaystyle \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\),

where (h, k) is the center of the ellipse, a is the horizontal radius, and b is the vertical radius. 

Plug in the coordinate pair:

\(\displaystyle \frac{(x-(-3))^2}{a^2} + \frac{(y-2)^2}{b^2} = 1\)

\(\displaystyle \frac{(x+3)^2}{a^2} + \frac{(y-2)^2}{b^2} = 1\)

Now we have to find the horizontal radius and the vertical radius. Let's compare points; we are told the ellipse passes through the point (-3, 6), which is vertically aligned with the center. Therefore the vertical radius is 4.

Similarly, the ellipse passes through the point (4, 2), which is horizontally aligned with the center. This means the horizontal radius must be 7.

Substitute:

\(\displaystyle \frac{(x+3)^2}{7^2} + \frac{(y-2)^2}{4^2} = 1\)

\(\displaystyle \frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1\)

Start by putting this equation in the standard form of the equation of an ellipse:

\(\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), where \(\displaystyle (h, k)\) is the center of the ellipse.

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 25x^2+4y^2-400x-8y+1504=0\)

\(\displaystyle 25x^2-400x+4y^2-8y+1504=0\)

Factor out \(\displaystyle 25\) from the \(\displaystyle x\) terms and \(\displaystyle 4\) from the \(\displaystyle y\) terms.

\(\displaystyle 25(x^2-16x)+4(y^2-2y)+1504=0\)

Now, complete the squares. Remember to add the same amount to both sides!

\(\displaystyle 25(x^2-16x+64)+4(y^2-2y+1)+1504=1604\)

Subtract \(\displaystyle 1504\) from both sides.

\(\displaystyle 25(x^2-16x+64)+4(y^2-2y+1)=100\)

Divide both sides by \(\displaystyle 100\).

\(\displaystyle \frac{(x^2-16x+64)}{4}+\frac{y^2-2y+1}{25}=1\)

Factor both terms to get the standard form of the equation of an ellipse.

\(\displaystyle \frac{(x-8)^2}{4}+\frac{(y+1)^2}{25}=1\)

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

Next, find the distance from the center to the focus of the ellipse, \(\displaystyle c\). Recall that when \(\displaystyle a>b\), the major axis will lie along the \(\displaystyle x\)-axis and be horizontal and that when \(\displaystyle b>a\), the major axis will lie along the \(\displaystyle y\)-axis and be vertical.

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{25-4}=\sqrt{21}\)

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because \(\displaystyle b>a\), the major axis for this ellipse is vertical. \(\displaystyle b\) will be the distance from the center to the vertices.

For this ellipse, \(\displaystyle b=5\).

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

\(\displaystyle e=\frac{\sqrt{21}}{5}\)

The general equation for an ellipse is \(\displaystyle \frac{(x-h)^2 }{a^2 } + \frac{(y-k)^2 }{b^2 } = 1\), although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has \(\displaystyle (h,k)\) as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at \(\displaystyle (1, 4)\) and the foci are at \(\displaystyle (1 \pm \sqrt{14} , 4 )\), we can see that the foci are \(\displaystyle \sqrt{14}\) away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so \(\displaystyle a = 7\). Since the foci are \(\displaystyle \sqrt{14}\) away from the center, we know that \(\displaystyle c = \sqrt{14}\). We can solve for b using the equation \(\displaystyle a^2 - c^2 = b^2\):

\(\displaystyle 7^2 - \sqrt{14}^2 = b^2\)

\(\displaystyle 49 - 14 = b^2\)

\(\displaystyle 35 = b^2\) that's really as far as we need to solve.

Putting all this information into the equation gives:

\(\displaystyle \frac{(x-1)^2 }{ 49} + \frac{(y - 4)^2 }{35 } = 1\)

Standard equation of an ellipse: \(\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\)

From the given information, the ellipse is centered around the origin (0,0), so h and k are both 0.

The coordinates of the vertices are on the x-axis, which is the major axis. The vertices are a units away from the center. Here, a = 10.

The coordinates of the co-vertices are on the y-axis, which is the minor axis. The co-vertices are b units away from the center. Here, b = 7.

\(\displaystyle \frac{x^{2}}{100}+\frac{y^{2}}{49}=1\)

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse. 

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

\(\displaystyle \frac{\left ( x- h\right )^{2}}{a^{2}} + \frac{\left ( y- k\right )^{2}}{b^{2}} = 1\)

as follows:

Separate the \(\displaystyle x\) and \(\displaystyle y\) terms:

\(\displaystyle 16x^{2} + 12y^{2} -96x-48y =0\)

\(\displaystyle (16x^{2} -96x) + (12y^{2}-48y) =0\)

Distribute out the quadratic coefficients:

\(\displaystyle 16 ( x^{2} -6x ) +12 ( y^{2}-4y) =0\)

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since \(\displaystyle \left (\frac{-6}{2} \right )^{2}= 9\) and \(\displaystyle \left (\frac{-4}{2} \right )^{2}= 4\), we get

\(\displaystyle 16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =0+ \textbf{?}+ \textbf{?}\)

Balance this equation, adjusting for the distributed coefficients:

\(\displaystyle 16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =16(9)+ 12(4)\)

\(\displaystyle 16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =192\)

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

\(\displaystyle 16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =192\)

Divide by 192:

\(\displaystyle \frac{16 ( x-3 )^{2}}{192} +\frac{12 ( y -2 ) ^{2} }{192}=\frac{192}{192}\)

\(\displaystyle \frac{( x-3 )^{2}}{12} +\frac{( y -2 ) ^{2} }{16}=1\)

The ellipse is now in standard form. \(\displaystyle b^{2} > a^{2}\), so the graph of the equation is a vertical ellipse 

The equation of the ellipse is given in the standard form

\(\displaystyle \frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}} = 1\)

This ellipse has its center at the origin \(\displaystyle (0,0)\). Also, since \(\displaystyle a^{2} >b^{2}\), it follows that the ellipse is horizontal. The foci are therefore along the horizontal axis of the ellipse; their coordinates are \(\displaystyle ( -c,0), (c,0)\), where

\(\displaystyle c = \sqrt{a^{2}-b^{2}}\)

Substituting 46 and 19 for \(\displaystyle a^{2}\) and \(\displaystyle b^{2}\), respectively,

\(\displaystyle c = \sqrt{46-19} = \sqrt{27} \approx 5.2\).

The foci are at the points \(\displaystyle (-5.2, 0 )\) and \(\displaystyle ( 5.2, 0 )\).  

This ellipse is in standard form

\(\displaystyle \frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}} = 1\)

where \(\displaystyle b^{2} > a^{2}\). This is a vertical ellipse, whose foci are 

\(\displaystyle c= \sqrt{b^{2}-a^{2}}\)

units from its center in a vertical direction.

The eccentricity of this  ellipse can be calculated by taking the ratio \(\displaystyle \frac{c}{b}\), or, equivalently, \(\displaystyle \frac{\sqrt{b^{2}-a^{2}}}{b}\). Set \(\displaystyle b^{2} = 169\) - making \(\displaystyle b= 13\) - and \(\displaystyle a^{2} = 25\). The eccentricity is calculated to be

\(\displaystyle \frac{\sqrt{169-25}}{13} =\frac{ \sqrt{144}}{13} = \frac{12}{13}\).