Recall 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.

When $\displaystyle a>b$, the major axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the major axis.

When $\displaystyle b>a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the major axis.

For the ellipse in question, $\displaystyle (0,0)$ is the center. In addition, $\displaystyle a=8$ and $\displaystyle b=7$. Since $\displaystyle a>b$, the major axis is horizontal and the endpoints are $\displaystyle (8, 0)$ and $\displaystyle (-8, 0)$

Recall 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.

When $\displaystyle a>b$, the major axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the major axis.

When $\displaystyle b>a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the major axis.

For the ellipse in question, $\displaystyle (0,0)$ is the center. In addition, $\displaystyle a=10$ and $\displaystyle b=20$. Since $\displaystyle b>a$, the major axis is vertical and the endpoints are $\displaystyle (0, 20)$ and $\displaystyle (0, -20)$.

Recall 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.

When $\displaystyle a>b$, the major axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the major axis.

When $\displaystyle b>a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the major axis.

For the ellipse in question, $\displaystyle (2,-11)$ is the center. In addition, $\displaystyle a=15$ and $\displaystyle b=16$. Since $\displaystyle b>a$, the major axis is vertical and the endpoints are $\displaystyle (2, -27)$ and $\displaystyle (2, 5)$.

Recall 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.

When $\displaystyle a>b$, the major axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the major axis.

When $\displaystyle b>a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the major axis.

For the ellipse in question, $\displaystyle (-5, 3)$ is the center. In addition, $\displaystyle a=13$ and $\displaystyle b=12$. Since $\displaystyle a>b$, the major axis is horizontal and the endpoints are $\displaystyle (8, 3)$ and $\displaystyle (-18, 3)$

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the $\displaystyle x$ terms and $\displaystyle y$ terms together.

$\displaystyle 25x^2+49y^2+200x-294y-284=0$

$\displaystyle 25x^2+200x+49y^2-294y-284=0$

Factor out $\displaystyle 25$ from the $\displaystyle x$ terms and $\displaystyle 49$ from the $\displaystyle y$ terms.

$\displaystyle 25(x^2-8x)+49(y^2-6y)-284=0$

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

$\displaystyle 25(x^2-8x+16)+49(y^2-6y+9)-284=841$

Add $\displaystyle 284$ to both sides.

$\displaystyle 25(x^2-8x+16)+49(y^2-6y+9)=1125$

Divide by $\displaystyle 1125$ on both sides.

$\displaystyle \frac{x^2-8x+16}{49}+\frac{y^2-6y+9}{25}=1$

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

$\displaystyle \frac{(x-4)^2}{49}+\frac{(y-3)^2}{25}=1$

Recall that when $\displaystyle a>b$, the major axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the major axis.

When $\displaystyle b>a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the major axis.

For the ellipse in question, $\displaystyle (4, 3)$ is the center. In addition, $\displaystyle a=7$ and $\displaystyle b=5$. Since $\displaystyle a>b$, the major axis is horizontal and the endpoints are $\displaystyle (11, 3)$ and $\displaystyle (-3, 3)$.

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the $\displaystyle x$ terms and $\displaystyle y$ terms together.

$\displaystyle 9x^2+18y^2-108x-360y+1962=0$

$\displaystyle 9x^2-108x+18y^2-360y+1962=0$

Factor out $\displaystyle 9$ from the $\displaystyle x$ terms and $\displaystyle 18$ from the $\displaystyle y$ terms.

$\displaystyle 9(x^2-12x)+18(y^2-20y)+1962=0$

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

$\displaystyle 9(x^2-12x+36)+18(y^2-20y+100)+1962=2124$

Subtract $\displaystyle 1962$ from both sides.

$\displaystyle 9(x^2-12x+36)+18(y^2-20y+100)=162$

Divide by $\displaystyle 162$ on both sides.

$\displaystyle \frac{x^2-12x+36}{18}+\frac{y^2-20x+100}{9}=1$

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

$\displaystyle \frac{(x-6)^2}{18}+\frac{(y-10)^2}{9}=1$

Recall that when $\displaystyle a>b$, the major axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the major axis.

When $\displaystyle b>a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the major axis.

For the ellipse in question, $\displaystyle (6, 10)$ is the center. In addition, $\displaystyle a=\sqrt{18}=3\sqrt2$ and $\displaystyle b=3$. Since $\displaystyle a>b$, the major axis is horizontal and the endpoints are $\displaystyle (6+3\sqrt2, 10)$ and $\displaystyle (6-3\sqrt2, 10)$.

Recall 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.

When $\displaystyle a< b$, the minor axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the minor axis.

When $\displaystyle b< a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the vertical minor axis.

For the ellipse in question, $\displaystyle (0,0)$ is the center. In addition, $\displaystyle a=10$ and $\displaystyle b=13$. Since $\displaystyle a< b$, the minor axis is horizontal and the endpoints are $\displaystyle (10, 0)$ and $\displaystyle (-10, 0)$

Recall 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.

When $\displaystyle a< b$, the minor axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the minor axis.

When $\displaystyle b< a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the vertical minor axis.

For the ellipse in question, $\displaystyle (0,0)$ is the center. In addition, $\displaystyle a=15$ and $\displaystyle b=20$. Since $\displaystyle a< b$, the minor axis is horizontal and the endpoints are $\displaystyle (15, 0)$ and $\displaystyle (-15, 0)$

Recall 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.

When $\displaystyle a< b$, the minor axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the minor axis.

When $\displaystyle b< a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the vertical minor axis.

For the ellipse in question, $\displaystyle (5,1)$ is the center. In addition, $\displaystyle a=5$ and $\displaystyle b=6$. Since $\displaystyle a< b$, the minor axis is horizontal and the endpoints are $\displaystyle (0,1)$ and $\displaystyle (10, 1)$.

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the $\displaystyle x$ terms and $\displaystyle y$ terms together.

$\displaystyle 25x^2+20y^2+200x-120y+80=0$

$\displaystyle 25x^2+200x+20y^2-120y+80=0$

Factor out $\displaystyle 25$ from the $\displaystyle x$ terms and $\displaystyle 20$ from the $\displaystyle y$ terms.

$\displaystyle 25(x^2+8x)+20(y^2-6y)+80=0$

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

$\displaystyle 25(x^2+8x+16)+20(y^2-6y+9)+80=580$

Subtract $\displaystyle 80$ from both sides.

$\displaystyle 25(x^2+8x+16)+20(y^2-6y+9)=500$

Divide by $\displaystyle 500$ on both sides.

$\displaystyle \frac{x^2+8x+16}{20}+\frac{y^2-6x+9}{25}=1$

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

$\displaystyle \frac{(x+4)^2}{20}+\frac{(y-3)^2}{25}=1$

When $\displaystyle a< b$, the minor axis is horizontal. In this case, $\displaystyle (h-a, k)$ and $\displaystyle (h+a, k)$ are the endpoints of the minor axis.

When $\displaystyle b< a$, $\displaystyle (h, k+b)$ and $\displaystyle (h, k-b)$ are the endpoints of the vertical minor axis.

For the ellipse in question, $\displaystyle (-4, 3)$ is the center. In addition, $\displaystyle a=\sqrt{20}=2\sqrt5$ and $\displaystyle b=5$. Since $\displaystyle a< b$, the minor axis is horizontal and the endpoints are $\displaystyle (-4+2\sqrt5, 3)$ and $\displaystyle (-4-2\sqrt5, 3)$.