Examples

(0, 0)

$(0, 0)$ ,

(4, 0)

$(4, 0)$ ,

(6, 0)

$(6, 0)$

Step 1

There are two general equations for an ellipse.

Horizontal ellipse equation

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

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

Vertical ellipse equation

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

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

Step 2

a

$a$ is the distance between the vertex

(6, 0)

$(6, 0)$ and the center point

(0, 0)

$(0, 0)$ .

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Step 2.1

Use the distance formula to determine the distance between the two points.

Distance = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 2.2

Substitute the actual values of the points into the distance formula.

a = \sqrt{{(6 - 0)}^{2} + {(0 - 0)}^{2}}

$a = \sqrt{{(6 - 0)}^{2} + {(0 - 0)}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Subtract

0

$0$ from

6

$6$ .

a = \sqrt{6^{2} + {(0 - 0)}^{2}}

$a = \sqrt{6^{2} + {(0 - 0)}^{2}}$

Step 2.3.2

Raise

6

$6$ to the power of

2

$2$ .

a = \sqrt{36 + {(0 - 0)}^{2}}

$a = \sqrt{36 + {(0 - 0)}^{2}}$

Step 2.3.3

Subtract

0

$0$ from

0

$0$ .

a = \sqrt{36 + 0^{2}}

$a = \sqrt{36 + 0^{2}}$

Step 2.3.4

Raising

0

$0$ to any positive power yields

0

$0$ .

a = \sqrt{36 + 0}

$a = \sqrt{36 + 0}$

Step 2.3.5

Add

36

$36$ and

0

$0$ .

a = \sqrt{36}

$a = \sqrt{36}$

Step 2.3.6

Rewrite

36

$36$ as

6^{2}

$6^{2}$ .

a = \sqrt{6^{2}}

$a = \sqrt{6^{2}}$

Step 2.3.7

Pull terms out from under the radical, assuming positive real numbers.

a = 6

$a = 6$

a = 6

$a = 6$

a = 6

$a = 6$

Step 3

c

$c$ is the distance between the focus

(4, 0)

$(4, 0)$ and the center

(0, 0)

$(0, 0)$ .

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Step 3.1

Use the distance formula to determine the distance between the two points.

Distance = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 3.2

Substitute the actual values of the points into the distance formula.

c = \sqrt{{(4 - 0)}^{2} + {(0 - 0)}^{2}}

$c = \sqrt{{(4 - 0)}^{2} + {(0 - 0)}^{2}}$

Step 3.3

Simplify.

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Step 3.3.1

Subtract

0

$0$ from

4

$4$ .

c = \sqrt{4^{2} + {(0 - 0)}^{2}}

$c = \sqrt{4^{2} + {(0 - 0)}^{2}}$

Step 3.3.2

Raise

4

$4$ to the power of

2

$2$ .

c = \sqrt{16 + {(0 - 0)}^{2}}

$c = \sqrt{16 + {(0 - 0)}^{2}}$

Step 3.3.3

Subtract

0

$0$ from

0

$0$ .

c = \sqrt{16 + 0^{2}}

$c = \sqrt{16 + 0^{2}}$

Step 3.3.4

Raising

0

$0$ to any positive power yields

0

$0$ .

c = \sqrt{16 + 0}

$c = \sqrt{16 + 0}$

Step 3.3.5

Add

16

$16$ and

0

$0$ .

c = \sqrt{16}

$c = \sqrt{16}$

Step 3.3.6

Rewrite

16

$16$ as

4^{2}

$4^{2}$ .

c = \sqrt{4^{2}}

$c = \sqrt{4^{2}}$

Step 3.3.7

Pull terms out from under the radical, assuming positive real numbers.

c = 4

$c = 4$

c = 4

$c = 4$

c = 4

$c = 4$

Step 4

Using the equation

c^{2} = a^{2} - b^{2}

$c^{2} = a^{2} - b^{2}$ . Substitute

6

$6$ for

a

$a$ and

4

$4$ for

c

$c$ .

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Step 4.1

Rewrite the equation as

{(6)}^{2} - b^{2} = 4^{2}

${(6)}^{2} - b^{2} = 4^{2}$ .

{(6)}^{2} - b^{2} = 4^{2}

${(6)}^{2} - b^{2} = 4^{2}$

Step 4.2

Raise

6

$6$ to the power of

2

$2$ .

36 - b^{2} = 4^{2}

$36 - b^{2} = 4^{2}$

Step 4.3

Raise

4

$4$ to the power of

2

$2$ .

36 - b^{2} = 16

$36 - b^{2} = 16$

Step 4.4

Move all terms not containing

b

$b$ to the right side of the equation.

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Step 4.4.1

Subtract

36

$36$ from both sides of the equation.

- b^{2} = 16 - 36

$- b^{2} = 16 - 36$

Step 4.4.2

Subtract

36

$36$ from

16

$16$ .

- b^{2} = - 20

$- b^{2} = - 20$

- b^{2} = - 20

$- b^{2} = - 20$

Step 4.5

Divide each term in

- b^{2} = - 20

$- b^{2} = - 20$ by

- 1

$- 1$ and simplify.

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Step 4.5.1

Divide each term in

- b^{2} = - 20

$- b^{2} = - 20$ by

- 1

$- 1$ .

\frac{- b^{2}}{- 1} = \frac{- 20}{- 1}

$\frac{- b^{2}}{- 1} = \frac{- 20}{- 1}$

Step 4.5.2

Simplify the left side.

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Step 4.5.2.1

Dividing two negative values results in a positive value.

\frac{b^{2}}{1} = \frac{- 20}{- 1}

$\frac{b^{2}}{1} = \frac{- 20}{- 1}$

Step 4.5.2.2

Divide

b^{2}

$b^{2}$ by

1

$1$ .

b^{2} = \frac{- 20}{- 1}

$b^{2} = \frac{- 20}{- 1}$

b^{2} = \frac{- 20}{- 1}

$b^{2} = \frac{- 20}{- 1}$

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Divide

- 20

$- 20$ by

- 1

$- 1$ .

b^{2} = 20

$b^{2} = 20$

b^{2} = 20

$b^{2} = 20$

b^{2} = 20

$b^{2} = 20$

Step 4.6

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

b = \pm \sqrt{20}

$b = \pm \sqrt{20}$

Step 4.7

Simplify

\pm \sqrt{20}

$\pm \sqrt{20}$ .

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Step 4.7.1

Rewrite

20

$20$ as

2^{2} \cdot 5

$2^{2} \cdot 5$ .

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Step 4.7.1.1

Factor

4

$4$ out of

20

$20$ .

b = \pm \sqrt{4 (5)}

$b = \pm \sqrt{4 (5)}$

Step 4.7.1.2

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

b = \pm \sqrt{2^{2} \cdot 5}

$b = \pm \sqrt{2^{2} \cdot 5}$

b = \pm \sqrt{2^{2} \cdot 5}

$b = \pm \sqrt{2^{2} \cdot 5}$

Step 4.7.2

Pull terms out from under the radical.

b = \pm 2 \sqrt{5}

$b = \pm 2 \sqrt{5}$

b = \pm 2 \sqrt{5}

$b = \pm 2 \sqrt{5}$

Step 4.8

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.8.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

b = 2 \sqrt{5}

$b = 2 \sqrt{5}$

Step 4.8.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

b = - 2 \sqrt{5}

$b = - 2 \sqrt{5}$

Step 4.8.3

The complete solution is the result of both the positive and negative portions of the solution.

b = 2 \sqrt{5}, - 2 \sqrt{5}

$b = 2 \sqrt{5}, - 2 \sqrt{5}$

b = 2 \sqrt{5}, - 2 \sqrt{5}

$b = 2 \sqrt{5}, - 2 \sqrt{5}$

b = 2 \sqrt{5}, - 2 \sqrt{5}

$b = 2 \sqrt{5}, - 2 \sqrt{5}$

Step 5

b

$b$ is a distance, which means it should be a positive number.

b = 2 \sqrt{5}

$b = 2 \sqrt{5}$

Step 6

The slope of the line between the focus

(4, 0)

$(4, 0)$ and the center

(0, 0)

$(0, 0)$ determines whether the ellipse is vertical or horizontal. If the slope is

0

$0$ , the graph is horizontal. If the slope is undefined, the graph is vertical.

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Step 6.1

Slope is equal to the change in

y

$y$ over the change in

x

$x$ , or rise over run.

m = \frac{change in y}{change in x}

$m = \frac{change in y}{change in x}$

Step 6.2

The change in

x

$x$ is equal to the difference in x-coordinates (also called run), and the change in

y

$y$ is equal to the difference in y-coordinates (also called rise).

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Step 6.3

Substitute in the values of

x

$x$ and

y

$y$ into the equation to find the slope.

m = \frac{0 - (0)}{0 - (4)}

$m = \frac{0 - (0)}{0 - (4)}$

Step 6.4

Simplify.

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Step 6.4.1

Simplify the numerator.

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Step 6.4.1.1

Multiply

- 1

$- 1$ by

0

$0$ .

m = \frac{0 + 0}{0 - (4)}

$m = \frac{0 + 0}{0 - (4)}$

Step 6.4.1.2

Add

0

$0$ and

0

$0$ .

m = \frac{0}{0 - (4)}

$m = \frac{0}{0 - (4)}$

m = \frac{0}{0 - (4)}

$m = \frac{0}{0 - (4)}$

Step 6.4.2

Simplify the denominator.

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Step 6.4.2.1

Multiply

- 1

$- 1$ by

4

$4$ .

m = \frac{0}{0 - 4}

$m = \frac{0}{0 - 4}$

Step 6.4.2.2

Subtract

4

$4$ from

0

$0$ .

m = \frac{0}{- 4}

$m = \frac{0}{- 4}$

m = \frac{0}{- 4}

$m = \frac{0}{- 4}$

Step 6.4.3

Divide

0

$0$ by

- 4

$- 4$ .

m = 0

$m = 0$

m = 0

$m = 0$

Step 6.5

The general equation for a horizontal ellipse is

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

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

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

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

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

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

Step 7

Substitute the values

h = 0

$h = 0$ ,

k = 0

$k = 0$ ,

a = 6

$a = 6$ , and

b = 2 \sqrt{5}

$b = 2 \sqrt{5}$ into

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

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the ellipse equation

\frac{{(x - (0))}^{2}}{{(6)}^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{{(x - (0))}^{2}}{{(6)}^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$ .

\frac{{(x - (0))}^{2}}{{(6)}^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{{(x - (0))}^{2}}{{(6)}^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8

Simplify to find the final equation of the ellipse.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{{(x + 0)}^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{{(x + 0)}^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.1.2

Add

x

$x$ and

0

$0$ .

\frac{x^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{x^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

\frac{x^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{x^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.2

Raise

6

$6$ to the power of

2

$2$ .

\frac{x^{2}}{36} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{x^{2}}{36} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.3

Simplify the numerator.

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Step 8.3.1

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{x^{2}}{36} + \frac{{(y + 0)}^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{x^{2}}{36} + \frac{{(y + 0)}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.3.2

Add

y

$y$ and

0

$0$ .

\frac{x^{2}}{36} + \frac{y^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{{(2 \sqrt{5})}^{2}} = 1$

\frac{x^{2}}{36} + \frac{y^{2}}{{(2 \sqrt{5})}^{2}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.4

Simplify the denominator.

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Step 8.4.1

Apply the product rule to

2 \sqrt{5}

$2 \sqrt{5}$ .

\frac{x^{2}}{36} + \frac{y^{2}}{2^{2} {\sqrt{5}}^{2}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{2^{2} {\sqrt{5}}^{2}} = 1$

Step 8.4.2

Raise

2

$2$ to the power of

2

$2$ .

\frac{x^{2}}{36} + \frac{y^{2}}{4 {\sqrt{5}}^{2}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{4 {\sqrt{5}}^{2}} = 1$

Step 8.4.3

Rewrite

{\sqrt{5}}^{2}

${\sqrt{5}}^{2}$ as

5

$5$ .

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Step 8.4.3.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{5}

$\sqrt{5}$ as

5^{\frac{1}{2}}

$5^{\frac{1}{2}}$ .

\frac{x^{2}}{36} + \frac{y^{2}}{4 {(5^{\frac{1}{2}})}^{2}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{4 {(5^{\frac{1}{2}})}^{2}} = 1$

Step 8.4.3.2

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{1}{2} \cdot 2}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{1}{2} \cdot 2}} = 1$

Step 8.4.3.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

2

$2$ .

\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{2}{2}}} = 1

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{2}{2}}} = 1$

Step 8.4.3.4

Cancel the common factor of

2

$2$ .

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Step 8.4.3.4.1

Cancel the common factor.

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{2}{2}}} = 1$

Step 8.4.3.4.2

Rewrite the expression.

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5} = 1$

Step 8.4.3.5

Evaluate the exponent.

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5} = 1$

Step 8.5

Multiply

$4$ by

$5$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$

Step 9

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