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 \(\displaystyle (-4-2\sqrt5, 3)\).