Examples

$(- 1, - 2)$ , $(2, - 2)$ , $(4, - 2)$

Step 1

There are two general equations for an ellipse.

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

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

Step 2

$a$ is the distance between the vertex $(4, - 2)$ and the center point $(- 1, - 2)$ .

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Step 2.1

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

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

Step 2.2

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

$a = \sqrt{{(4 - (- 1))}^{2} + {((- 2) - (- 2))}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

$a = \sqrt{{(4 + 1)}^{2} + {((- 2) - (- 2))}^{2}}$

Step 2.3.2

Add $4$ and $1$ .

$a = \sqrt{5^{2} + {((- 2) - (- 2))}^{2}}$

Step 2.3.3

Raise $5$ to the power of $2$ .

$a = \sqrt{25 + {((- 2) - (- 2))}^{2}}$

Step 2.3.4

Multiply $- 1$ by $- 2$ .

$a = \sqrt{25 + {(- 2 + 2)}^{2}}$

Step 2.3.5

Add $- 2$ and $2$ .

$a = \sqrt{25 + 0^{2}}$

Step 2.3.6

Raising $0$ to any positive power yields $0$ .

$a = \sqrt{25 + 0}$

Step 2.3.7

Add $25$ and $0$ .

$a = \sqrt{25}$

Step 2.3.8

Rewrite $25$ as $5^{2}$ .

$a = \sqrt{5^{2}}$

Step 2.3.9

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

$a = 5$

Step 3

$c$ is the distance between the focus $(2, - 2)$ and the center $(- 1, - 2)$ .

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Step 3.1

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

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

Step 3.2

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

$c = \sqrt{{(2 - (- 1))}^{2} + {((- 2) - (- 2))}^{2}}$

Step 3.3

Simplify.

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Step 3.3.1

Multiply $- 1$ by $- 1$ .

$c = \sqrt{{(2 + 1)}^{2} + {((- 2) - (- 2))}^{2}}$

Step 3.3.2

Add $2$ and $1$ .

$c = \sqrt{3^{2} + {((- 2) - (- 2))}^{2}}$

Step 3.3.3

Raise $3$ to the power of $2$ .

$c = \sqrt{9 + {((- 2) - (- 2))}^{2}}$

Step 3.3.4

Multiply $- 1$ by $- 2$ .

$c = \sqrt{9 + {(- 2 + 2)}^{2}}$

Step 3.3.5

Add $- 2$ and $2$ .

$c = \sqrt{9 + 0^{2}}$

Step 3.3.6

Raising $0$ to any positive power yields $0$ .

$c = \sqrt{9 + 0}$

Step 3.3.7

Add $9$ and $0$ .

$c = \sqrt{9}$

Step 3.3.8

Rewrite $9$ as $3^{2}$ .

$c = \sqrt{3^{2}}$

Step 3.3.9

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

$c = 3$

Step 4

Using the equation $c^{2} = a^{2} - b^{2}$ . Substitute $5$ for $a$ and $3$ for $c$ .

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Step 4.1

Rewrite the equation as ${(5)}^{2} - b^{2} = 3^{2}$ .

${(5)}^{2} - b^{2} = 3^{2}$

Step 4.2

Raise $5$ to the power of $2$ .

$25 - b^{2} = 3^{2}$

Step 4.3

Raise $3$ to the power of $2$ .

$25 - b^{2} = 9$

Step 4.4

Move all terms not containing $b$ to the right side of the equation.

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Step 4.4.1

Subtract $25$ from both sides of the equation.

$- b^{2} = 9 - 25$

Step 4.4.2

Subtract $25$ from $9$ .

$- b^{2} = - 16$

Step 4.5

Divide each term in $- b^{2} = - 16$ by $- 1$ and simplify.

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Step 4.5.1

Divide each term in $- b^{2} = - 16$ by $- 1$ .

$\frac{- b^{2}}{- 1} = \frac{- 16}{- 1}$

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.

$\frac{b^{2}}{1} = \frac{- 16}{- 1}$

Step 4.5.2.2

Divide $b^{2}$ by $1$ .

$b^{2} = \frac{- 16}{- 1}$

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Divide $- 16$ by $- 1$ .

$b^{2} = 16$

Step 4.6

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

$b = \pm \sqrt{16}$

Step 4.7

Simplify $\pm \sqrt{16}$ .

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Step 4.7.1

Rewrite $16$ as $4^{2}$ .

$b = \pm \sqrt{4^{2}}$

Step 4.7.2

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

$b = \pm 4$

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 $\pm$ to find the first solution.

$b = 4$

Step 4.8.2

Next, use the negative value of the $\pm$ to find the second solution.

$b = - 4$

Step 4.8.3

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

$b = 4, - 4$

Step 5

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

$b = 4$

Step 6

The slope of the line between the focus $(2, - 2)$ and the center $(- 1, - 2)$ determines whether the ellipse is vertical or horizontal. If the slope is $0$ , 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 $y$ over the change in $x$ , or rise over run.

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

Step 6.2

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

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

Step 6.3

Substitute in the values of $x$ and $y$ into the equation to find the slope.

$m = \frac{- 2 - (- 2)}{- 1 - (2)}$

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 $- 1$ by $- 2$ .

$m = \frac{- 2 + 2}{- 1 - (2)}$

Step 6.4.1.2

Add $- 2$ and $2$ .

$m = \frac{0}{- 1 - (2)}$

Step 6.4.2

Simplify the denominator.

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Step 6.4.2.1

Multiply $- 1$ by $2$ .

$m = \frac{0}{- 1 - 2}$

Step 6.4.2.2

Subtract $2$ from $- 1$ .

$m = \frac{0}{- 3}$

Step 6.4.3

Divide $0$ by $- 3$ .

$m = 0$

Step 6.5

The general equation for a horizontal ellipse is $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ .

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

Step 7

Substitute the values $h = - 1$ , $k = - 2$ , $a = 5$ , and $b = 4$ into $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the ellipse equation $\frac{{(x - (- 1))}^{2}}{{(5)}^{2}} + \frac{{(y - (- 2))}^{2}}{{(4)}^{2}} = 1$ .

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

Step 8

Simplify to find the final equation of the ellipse.

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Step 8.1

Multiply $- 1$ by $- 1$ .

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

Step 8.2

Raise $5$ to the power of $2$ .

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

Step 8.3

Multiply $- 1$ by $- 2$ .

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

Step 8.4

Raise $4$ to the power of $2$ .

$\frac{{(x + 1)}^{2}}{25} + \frac{{(y + 2)}^{2}}{16} = 1$

Step 9

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