Precalculus Examples

vertices at

(6, 0)

$(6, 0)$ ,

(- 6, 0)

$(- 6, 0)$ ; foci at

(8, 0)

$(8, 0)$ ,

(- 8, 0)

$(- 8, 0)$

Step 1

Find the center

(h, k)

$(h, k)$ by finding the midpoint of the given vertices.

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Step 1.1

Use the midpoint formula to find the midpoint of the line segment.

(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Step 1.2

Substitute in the values for

(x_{1}, y_{1})

$(x_{1}, y_{1})$ and

(x_{2}, y_{2})

$(x_{2}, y_{2})$ .

(\frac{- 6 + 6}{2}, \frac{0 + 0}{2})

$(\frac{- 6 + 6}{2}, \frac{0 + 0}{2})$

Step 1.3

Cancel the common factor of

- 6 + 6

$- 6 + 6$ and

2

$2$ .

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Step 1.3.1

Factor

2

$2$ out of

- 6

$- 6$ .

(\frac{2 \cdot - 3 + 6}{2}, \frac{0 + 0}{2})

$(\frac{2 \cdot - 3 + 6}{2}, \frac{0 + 0}{2})$

Step 1.3.2

Factor

2

$2$ out of

6

$6$ .

(\frac{2 \cdot - 3 + 2 \cdot 3}{2}, \frac{0 + 0}{2})

$(\frac{2 \cdot - 3 + 2 \cdot 3}{2}, \frac{0 + 0}{2})$

Step 1.3.3

Factor

2

$2$ out of

2 \cdot - 3 + 2 \cdot 3

$2 \cdot - 3 + 2 \cdot 3$ .

(\frac{2 \cdot (- 3 + 3)}{2}, \frac{0 + 0}{2})

$(\frac{2 \cdot (- 3 + 3)}{2}, \frac{0 + 0}{2})$

Step 1.3.4

Cancel the common factors.

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Step 1.3.4.1

Factor

2

$2$ out of

2

$2$ .

(\frac{2 \cdot (- 3 + 3)}{2 (1)}, \frac{0 + 0}{2})

$(\frac{2 \cdot (- 3 + 3)}{2 (1)}, \frac{0 + 0}{2})$

Step 1.3.4.2

Cancel the common factor.

$(\frac{2 \cdot (- 3 + 3)}{2 \cdot 1}, \frac{0 + 0}{2})$

Step 1.3.4.3

Rewrite the expression.

$(\frac{- 3 + 3}{1}, \frac{0 + 0}{2})$

Step 1.3.4.4

Divide

$- 3 + 3$ by

$1$ .

$(- 3 + 3, \frac{0 + 0}{2})$

Step 1.4

Add

$- 3$ and

$3$ .

$(0, \frac{0 + 0}{2})$

Step 1.5

Cancel the common factor of

$0 + 0$ and

$2$ .

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Step 1.5.1

Factor

$2$ out of

$0$ .

$(0, \frac{2 (0) + 0}{2})$

Step 1.5.2

Factor

$2$ out of

$0$ .

$(0, \frac{2 \cdot 0 + 2 \cdot 0}{2})$

Step 1.5.3

Factor

$2$ out of

$2 \cdot 0 + 2 \cdot 0$ .

$(0, \frac{2 \cdot (0 + 0)}{2})$

Step 1.5.4

Cancel the common factors.

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Step 1.5.4.1

Factor

$2$ out of

$2$ .

$(0, \frac{2 \cdot (0 + 0)}{2 (1)})$

Step 1.5.4.2

Cancel the common factor.

$(0, \frac{2 \cdot (0 + 0)}{2 \cdot 1})$

Step 1.5.4.3

Rewrite the expression.

$(0, \frac{0 + 0}{1})$

Step 1.5.4.4

Divide

$0 + 0$ by

$1$ .

$(0, 0 + 0)$

Step 1.6

Add

$0$ and

$0$ .

$(0, 0)$

Step 2

Graph the center and the given foci and vertices. Because the points lie horizontally, the hyperbola opens to the left and right and the formula of the hyperbola will be

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

Step 3

Find

$a$ by finding the distance between a vertex and the center point.

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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.

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

Step 3.3

Simplify.

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Step 3.3.1

Subtract

$0$ from

$- 6$ .

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

Step 3.3.2

Raise

$- 6$ to the power of

$2$ .

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

Step 3.3.3

Subtract

$0$ from

$0$ .

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

Step 3.3.4

Raising

$0$ to any positive power yields

$0$ .

$a = \sqrt{36 + 0}$

Step 3.3.5

Add

$36$ and

$0$ .

$a = \sqrt{36}$

Step 3.3.6

Rewrite

$36$ as

$6^{2}$ .

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

Step 3.3.7

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

$a = 6$

Step 4

Find

$c$ by finding the distance between a focus and the center point.

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Step 4.1

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 4.2

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

$c = \sqrt{{((- 8) - 0)}^{2} + {(0 - 0)}^{2}}$

Step 4.3

Simplify.

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Step 4.3.1

Subtract

$0$ from

$- 8$ .

$c = \sqrt{{(- 8)}^{2} + {(0 - 0)}^{2}}$

Step 4.3.2

Raise

$- 8$ to the power of

$2$ .

$c = \sqrt{64 + {(0 - 0)}^{2}}$

Step 4.3.3

Subtract

$0$ from

$0$ .

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

Step 4.3.4

Raising

$0$ to any positive power yields

$0$ .

$c = \sqrt{64 + 0}$

Step 4.3.5

Add

$64$ and

$0$ .

$c = \sqrt{64}$

Step 4.3.6

Rewrite

$64$ as

$8^{2}$ .

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

Step 4.3.7

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

$c = 8$

Step 5

Plug the values of

$a$ and

$c$ into

$c^{2} = a^{2} + b^{2}$ and solve for

$b^{2}$ .

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Step 5.1

Plug in

$8$ for

$c$ and

$6$ for

$a$ .

$8^{2} = 6^{2} + b^{2}$

Step 5.2

Rewrite the equation as

$6^{2} + b^{2} = 8^{2}$ .

$6^{2} + b^{2} = 8^{2}$

Step 5.3

Raise

$6$ to the power of

$2$ .

$36 + b^{2} = 8^{2}$

Step 5.4

Raise

$8$ to the power of

$2$ .

$36 + b^{2} = 64$

Step 5.5

Move all terms not containing

$b^{2}$ to the right side of the equation.

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Step 5.5.1

Subtract

$36$ from both sides of the equation.

$b^{2} = 64 - 36$

Step 5.5.2

Subtract

$36$ from

$64$ .

$b^{2} = 28$

Step 6

Substitute the found values into the formula and simplify.

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Step 6.1

Substitute the found values into the formula.

$\frac{{(x + 0)}^{2}}{6^{2}} - \frac{{(y + 0)}^{2}}{28} = 1$

Step 6.2

Combine the opposite terms in

$\frac{{(x + 0)}^{2}}{6^{2}} - \frac{{(y + 0)}^{2}}{28}$ .

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Step 6.2.1

Add

$x$ and

$0$ .

$\frac{x^{2}}{6^{2}} - \frac{{(y + 0)}^{2}}{28} = 1$

Step 6.2.2

Add

$y$ and

$0$ .

$\frac{x^{2}}{6^{2}} - \frac{y^{2}}{28} = 1$

Step 6.3

Raise

$6$ to the power of

$2$ .

$\frac{x^{2}}{36} - \frac{y^{2}}{28} = 1$

Step 7