Start with the equation

$\displaystyle 9x^2 - 18x + 100y^2 + 600y = -9$

It would be helpful if we "complete the square" for both the x and y polynomials.

• $\displaystyle (9x^2- 18x) + 9=9(x - 1)^2$
• $\displaystyle (100y^2 + 600y) + 900 = 100(y^2+6x +9) = 100(y+ 3)2$

So by adding 9 and 900 we transform the equation into

• $\displaystyle 9x^2 - 18x + 9 + 100y^2 + 600y + 900 = -9 + 9 + 900$
• $\displaystyle 9(x-1)^2 + 100(y+3)^2 = 900$

And then by dividing 900, we see that the outcome is:

$\displaystyle \frac{(x-1)^2}{100} + \frac{(y+3)^2}{9} = 1$

The standard equation for an ellipse is

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

With a center at $\displaystyle (h, k)$

Therefore this is the equation of an ellipse, with center $\displaystyle (1, -3)$

Remember that the equation for an ellipse in standard form looks like the following:

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

Where the point (h,k) gives the center of the ellipse, a is half the length of its axis in the x direction, and b is half the length of its axis in the y direction. We can see that this form has a 1 on the right side of the equation, so let's start by dividing both sides of our equation by 36 to get a 1 on the right side:

$\displaystyle 4(x-4)^2+9(x-7)^2=36$

$\displaystyle \frac{4(x-4)^2+9(x-7)^2}{36}=\frac{36}{36}$

$\displaystyle \frac{4(x-4)^2}{36}+\frac{9(x-7)^2}{36}=1$

Now we can simplify the fractions on the right side of the equation, which gives us the equation for our ellipse in standard form:

$\displaystyle \frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

This ellipse would have its center at (4,7), would be 6 units wide in the x direction, and 4 units wide in the y direction, because $\displaystyle a^2=9$ so $\displaystyle a=3$, and $\displaystyle b^2=4$ so $\displaystyle b=2$.

Remember that in order for the equation of an ellipse to be written in standard form, it must be written in one of the following two ways:

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

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

Where the point (h,k) gives the center of the ellipse, a is half the length of the axis for which it is the denominator, and b is half the length of the axis for which it is the denominator. Looking at our answer choices, we need one that involves addition, has different denominators under the x and y terms, and is equal to 1 on the right side of the equation. We can see that out of all the answer choices, the following is the only one that satisfies those requirements of standard form:

$\displaystyle \frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

The equation of an ellipse in standard form is

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

where $\displaystyle (h,k)$ is the center point, $\displaystyle a$ is half the length of its axis in the $\displaystyle x$ direction, and $\displaystyle b$ is half the length of the axis in the $\displaystyle y$ direction.

In this instance, the center point $\displaystyle (h,k)=$ $\displaystyle (3,6)$, $\displaystyle a=\frac{1}2{(18)}=9$, and $\displaystyle b=\frac{1}{2}(10)=5$. Therefore:

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

$\displaystyle \frac{(x-3)^{2}}{9^{2}}+\frac{(y-6)^{2}}{5^{2}}=1$

$\displaystyle \frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

Recall that the general equation of a circle is: 

$\displaystyle (x - h)^2 + (y - k)^2 = r^2$. 

The key feature here being that both the $\displaystyle x$ and $\displaystyle y$ terms are squared.

Thus, we can narrow down the choices without this feature.

Next, consider the center. Since the center here is $\displaystyle (0,0)$, we should not have any middle terms.

The only choice that fits that description is $\displaystyle x^2 + y^2 = 144$. 

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:

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

The formula for an ellipse centered at the point $\displaystyle (h, k)$ with a horizontal major axis (ie: parallel to the x-axis) has the formula

$\displaystyle \small \frac{(x-h)^2}{[\frac{a}{2}]^2}+\frac{(y-k)^2}{[\frac{b}{2}]^2}=1$ 

where $\displaystyle \small a$ and $\displaystyle \small b$ are the lengths of the major and minor axes respectively.

Since the origin is at $\displaystyle \small (0,0)$ and $\displaystyle \small a=8$ and $\displaystyle \small b=4$ in this problem, the equation is

$\displaystyle \small \frac{(x-0)^2}{[\frac{8}{2}]^2}+\frac{(y-0)^2}{[\frac{4}{2}]^2}=1$

or

$\displaystyle \small \frac{x^2}{16}+\frac{y^2}{4}=1$

Because our equation is already in the format

 $\displaystyle \frac{(x-h)^2}{w^2}+\frac{(y-k)^2}{v^2}=1$,

we do not have to manipulate the equation. The center of any ellipse in this form will always be $\displaystyle (h,k)$. So in this case, our center will be $\displaystyle (-3,7)$. To find the foci of the ellipse, we must use the equation $\displaystyle b^2+c^2=a^2$, where $\displaystyle a^2$ is the greater of the two denominators in our equation ($\displaystyle w^2$ and $\displaystyle v^2$), $\displaystyle b^2$ is the lesser and $\displaystyle c$ is the distance from the center to the foci.

We know that $\displaystyle a^2=36$ and $\displaystyle b^2=9$.

By using $\displaystyle a^2-b^2=c^2$, we see that $\displaystyle c^2=25$, so $\displaystyle c=5$.

We now know that the two foci are going to be $\displaystyle 5$ units in either direction of the center along the greater axis.

Because the greater denominator is under our term containing $\displaystyle y$, our ellipse will have its greater axis going vertically, rather than horizontally. Therefore, our foci will be $\displaystyle 5$ units above and below our center, at $\displaystyle (-3,2)$ and $\displaystyle (-3,12)$.

Recall that the standard form of the equation of an ellipse is

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

For the equation given in the question, $\displaystyle h=1$ and $\displaystyle k=-2$. 

The center of the ellipse is at $\displaystyle (1, -2)$

Recall that the standard form of the equation of an ellipse is

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

For the equation given in the question, $\displaystyle h=-9$ and $\displaystyle k=4$. 

The center of the ellipse is at $\displaystyle (-9, 4)$.