Examples

$(- 1, 2)$ , $(5, 2)$ , $(7, 2)$

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 $(7, 2)$ 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.

$a = \sqrt{{(7 - (- 1))}^{2} + {(2 - 2)}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

$a = \sqrt{{(7 + 1)}^{2} + {(2 - 2)}^{2}}$

Step 2.3.2

Add $7$ and $1$ .

$a = \sqrt{8^{2} + {(2 - 2)}^{2}}$

Step 2.3.3

Raise $8$ to the power of $2$ .

$a = \sqrt{64 + {(2 - 2)}^{2}}$

Step 2.3.4

Subtract $2$ from $2$ .

$a = \sqrt{64 + 0^{2}}$

Step 2.3.5

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

$a = \sqrt{64 + 0}$

Step 2.3.6

Add $64$ and $0$ .

$a = \sqrt{64}$

Step 2.3.7

Rewrite $64$ as $8^{2}$ .

$a = \sqrt{8^{2}}$

Step 2.3.8

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

$a = 8$

Step 3

$c$ is the distance between the focus $(5, 2)$ 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.

$c = \sqrt{{(5 - (- 1))}^{2} + {(2 - 2)}^{2}}$

Step 3.3

Simplify.

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Step 3.3.1

Multiply $- 1$ by $- 1$ .

$c = \sqrt{{(5 + 1)}^{2} + {(2 - 2)}^{2}}$

Step 3.3.2

Add $5$ and $1$ .

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

Step 3.3.3

Raise $6$ to the power of $2$ .

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

Step 3.3.4

Subtract $2$ from $2$ .

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

Step 3.3.5

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

$c = \sqrt{36 + 0}$

Step 3.3.6

Add $36$ and $0$ .

$c = \sqrt{36}$

Step 3.3.7

Rewrite $36$ as $6^{2}$ .

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

Step 3.3.8

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

$c = 6$

Step 4

Using the equation $c^{2} = a^{2} - b^{2}$ . Substitute $8$ for $a$ and $6$ for $c$ .

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Step 4.1

Rewrite the equation as ${(8)}^{2} - b^{2} = 6^{2}$ .

${(8)}^{2} - b^{2} = 6^{2}$

Step 4.2

Raise $8$ to the power of $2$ .

$64 - b^{2} = 6^{2}$

Step 4.3

Raise $6$ to the power of $2$ .

$64 - b^{2} = 36$

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 $64$ from both sides of the equation.

$- b^{2} = 36 - 64$

Step 4.4.2

Subtract $64$ from $36$ .

$- b^{2} = - 28$

Step 4.5

Divide each term in $- b^{2} = - 28$ by $- 1$ and simplify.

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Step 4.5.1

Divide each term in $- b^{2} = - 28$ by $- 1$ .

$\frac{- b^{2}}{- 1} = \frac{- 28}{- 1}$

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.

$\frac{b^{2}}{1} = \frac{- 28}{- 1}$

Step 4.5.2.2

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

$b^{2} = \frac{- 28}{- 1}$

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Divide $- 28$ by $- 1$ .

$b^{2} = 28$

Step 4.6

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

$b = \pm \sqrt{28}$

Step 4.7

Simplify $\pm \sqrt{28}$ .

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Step 4.7.1

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 4.7.1.1

Factor $4$ out of $28$ .

$b = \pm \sqrt{4 (7)}$

Step 4.7.1.2

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

$b = \pm \sqrt{2^{2} \cdot 7}$

Step 4.7.2

Pull terms out from under the radical.

$b = \pm 2 \sqrt{7}$

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.

$b = 2 \sqrt{7}$

Step 4.8.2

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

$b = - 2 \sqrt{7}$

Step 4.8.3

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

$b = 2 \sqrt{7}, - 2 \sqrt{7}$

Step 5

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

$b = 2 \sqrt{7}$

Step 6

The slope of the line between the focus $(5, 2)$ 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.

$m = \frac{2 - (2)}{- 1 - (5)}$

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 $2$ .

$m = \frac{2 - 2}{- 1 - (5)}$

Step 6.4.1.2

Subtract $2$ from $2$ .

$m = \frac{0}{- 1 - (5)}$

Step 6.4.2

Simplify the denominator.

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Step 6.4.2.1

Multiply $- 1$ by $5$ .

$m = \frac{0}{- 1 - 5}$

Step 6.4.2.2

Subtract $5$ from $- 1$ .

$m = \frac{0}{- 6}$

Step 6.4.3

Divide $0$ by $- 6$ .

$m = 0$

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 $h = - 1$ , $k = 2$ , $a = 8$ , and $b = 2 \sqrt{7}$ into $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the ellipse equation $\frac{{(x - (- 1))}^{2}}{{(8)}^{2}} + \frac{{(y - (2))}^{2}}{{(2 \sqrt{7})}^{2}} = 1$ .

$\frac{{(x - (- 1))}^{2}}{{(8)}^{2}} + \frac{{(y - (2))}^{2}}{{(2 \sqrt{7})}^{2}} = 1$

Step 8

Simplify to find the final equation of the ellipse.

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Step 8.1

Multiply $- 1$ by $- 1$ .

$\frac{{(x + 1)}^{2}}{8^{2}} + \frac{{(y - (2))}^{2}}{{(2 \sqrt{7})}^{2}} = 1$

Step 8.2

Raise $8$ to the power of $2$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - (2))}^{2}}{{(2 \sqrt{7})}^{2}} = 1$

Step 8.3

Multiply $- 1$ by $2$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{{(2 \sqrt{7})}^{2}} = 1$

Step 8.4

Simplify the denominator.

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Step 8.4.1

Apply the product rule to $2 \sqrt{7}$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{2^{2} {\sqrt{7}}^{2}} = 1$

Step 8.4.2

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

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 {\sqrt{7}}^{2}} = 1$

Step 8.4.3

Rewrite ${\sqrt{7}}^{2}$ as $7$ .

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Step 8.4.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7}$ as $7^{\frac{1}{2}}$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 {(7^{\frac{1}{2}})}^{2}} = 1$

Step 8.4.3.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 \cdot 7^{\frac{1}{2} \cdot 2}} = 1$

Step 8.4.3.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 \cdot 7^{\frac{2}{2}}} = 1$

Step 8.4.3.4

Cancel the common factor of $2$ .

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Step 8.4.3.4.1

Cancel the common factor.

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 \cdot 7^{\frac{2}{2}}} = 1$

Step 8.4.3.4.2

Rewrite the expression.

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 \cdot 7} = 1$

Step 8.4.3.5

Evaluate the exponent.

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{4 \cdot 7} = 1$

Step 8.5

Multiply $4$ by $7$ .

$\frac{{(x + 1)}^{2}}{64} + \frac{{(y - 2)}^{2}}{28} = 1$

Step 9

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