Algebra Examples

(1, 2)

$(1, 2)$ ,

(4, 2)

$(4, 2)$ ,

(5, 2)

$(5, 2)$

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

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

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

(5, 2)

$(5, 2)$ and the center point

(1, 2)

$(1, 2)$ .

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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}^{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{{(5 - 1)}^{2} + {(2 - 2)}^{2}}

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

Step 2.3

Simplify.

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Step 2.3.1

Subtract

1

$1$ from

5

$5$ .

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

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

Step 2.3.2

Raise

$4$ to the power of

$2$ .

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

Step 2.3.3

Subtract

$2$ from

$2$ .

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

Step 2.3.4

Raising

$0$ to any positive power yields

$0$ .

$a = \sqrt{16 + 0}$

Step 2.3.5

Add

$16$ and

$0$ .

$a = \sqrt{16}$

Step 2.3.6

Rewrite

$16$ as

$4^{2}$ .

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

Step 2.3.7

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

$a = 4$

Step 3

$c$ is the distance between the focus

$(4, 2)$ 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.

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

Step 3.3

Simplify.

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Step 3.3.1

Subtract

$1$ from

$4$ .

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

Step 3.3.2

Raise

$3$ to the power of

$2$ .

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

Step 3.3.3

Subtract

$2$ from

$2$ .

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

Step 3.3.4

Raising

$0$ to any positive power yields

$0$ .

$c = \sqrt{9 + 0}$

Step 3.3.5

Add

$9$ and

$0$ .

$c = \sqrt{9}$

Step 3.3.6

Rewrite

$9$ as

$3^{2}$ .

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

Step 3.3.7

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

$4$ for

$a$ and

$3$ for

$c$ .

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Step 4.1

Rewrite the equation as

${(4)}^{2} - b^{2} = 3^{2}$ .

${(4)}^{2} - b^{2} = 3^{2}$

Step 4.2

Raise

$4$ to the power of

$2$ .

$16 - b^{2} = 3^{2}$

Step 4.3

Raise

$3$ to the power of

$2$ .

$16 - b^{2} = 9$

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

$16$ from both sides of the equation.

$- b^{2} = 9 - 16$

Step 4.4.2

Subtract

$16$ from

$9$ .

$- b^{2} = - 7$

Step 4.5

Divide each term in

$- b^{2} = - 7$ by

$- 1$ and simplify.

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Step 4.5.1

Divide each term in

$- b^{2} = - 7$ by

$- 1$ .

$\frac{- b^{2}}{- 1} = \frac{- 7}{- 1}$

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.

$\frac{b^{2}}{1} = \frac{- 7}{- 1}$

Step 4.5.2.2

Divide

$b^{2}$ by

$1$ .

$b^{2} = \frac{- 7}{- 1}$

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Divide

$- 7$ by

$- 1$ .

$b^{2} = 7$

Step 4.6

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

$b = \pm \sqrt{7}$

Step 4.7

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

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Step 4.7.1

First, use the positive value of the

$\pm$ to find the first solution.

$b = \sqrt{7}$

Step 4.7.2

Next, use the negative value of the

$\pm$ to find the second solution.

$b = - \sqrt{7}$

Step 4.7.3

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

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

Step 5

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

$b = \sqrt{7}$

Step 6

The slope of the line between the focus

$(4, 2)$ 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.

$m = \frac{2 - (2)}{1 - (4)}$

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

$- 1$ by

$2$ .

$m = \frac{2 - 2}{1 - (4)}$

Step 6.4.1.2

Subtract

$2$ from

$2$ .

$m = \frac{0}{1 - (4)}$

Step 6.4.2

Simplify the denominator.

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Step 6.4.2.1

Multiply

$- 1$ by

$4$ .

$m = \frac{0}{1 - 4}$

Step 6.4.2.2

Subtract

$4$ 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

$h = 1$ ,

$k = 2$ ,

$a = 4$ , and

$b = \sqrt{7}$ into

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

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

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

Step 8

Simplify to find the final equation of the ellipse.

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Step 8.1

Multiply

$- 1$ by

$1$ .

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

Step 8.2

Raise

$4$ to the power of

$2$ .

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - (2))}^{2}}{{(\sqrt{7})}^{2}} = 1$

Step 8.3

Multiply

$- 1$ by

$2$ .

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{{\sqrt{7}}^{2}} = 1$

Step 8.4

Rewrite

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

$7$ .

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Step 8.4.1

Use

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

$\sqrt{7}$ as

$7^{\frac{1}{2}}$ .

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{{(7^{\frac{1}{2}})}^{2}} = 1$

Step 8.4.2

Apply the power rule and multiply exponents,

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

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{7^{\frac{1}{2} \cdot 2}} = 1$

Step 8.4.3

Combine

$\frac{1}{2}$ and

$2$ .

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{7^{\frac{2}{2}}} = 1$

Step 8.4.4

Cancel the common factor of

$2$ .

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Step 8.4.4.1

Cancel the common factor.

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{7^{\frac{2}{2}}} = 1$

Step 8.4.4.2

Rewrite the expression.

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{7} = 1$

Step 8.4.5

Evaluate the exponent.

$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 2)}^{2}}{7} = 1$

Step 9

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