Algebra Examples

$(2, 3)$ , $(1, 3)$ , $(- 4, 3)$

There are two general equations for a hyperbola.

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

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

$a$ is the distance between the vertex $(1, 3)$ and the center point $(2, 3)$ .

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

$a = D i s t a n c e = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

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

$a = \sqrt{{((1) - (2))}^{2} + {((3) - (3))}^{2}}$

Simplify.

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Multiply $- 1$ by $2$ .

$a = \sqrt{{(1 - 2)}^{2} + {((3) - (3))}^{2}}$

Subtract $2$ from $1$ .

$a = \sqrt{{(- 1)}^{2} + {((3) - (3))}^{2}}$

Raise $- 1$ to the power of $2$ .

$a = \sqrt{1 + {((3) - (3))}^{2}}$

Multiply $- 1$ by $3$ .

$a = \sqrt{1 + {(3 - 3)}^{2}}$

Subtract $3$ from $3$ .

$a = \sqrt{1 + {(0)}^{2}}$

Remove parentheses around $0$ .

$a = \sqrt{1 + 0^{2}}$

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

$a = \sqrt{1 + 0}$

Add $1$ and $0$ .

$a = \sqrt{1}$

Any root of $1$ is $1$ .

$a = 1$

$c$ is the distance between the focus $(- 4, 3)$ and the center $(2, 3)$ .

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

$c = D i s t a n c e = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

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

$c = \sqrt{{((- 4) - (2))}^{2} + {((3) - (3))}^{2}}$

Simplify.

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Multiply $- 1$ by $2$ .

$c = \sqrt{{(- 4 - 2)}^{2} + {((3) - (3))}^{2}}$

Subtract $2$ from $- 4$ .

$c = \sqrt{{(- 6)}^{2} + {((3) - (3))}^{2}}$

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

$c = \sqrt{36 + {((3) - (3))}^{2}}$

Multiply $- 1$ by $3$ .

$c = \sqrt{36 + {(3 - 3)}^{2}}$

Subtract $3$ from $3$ .

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

Remove parentheses around $0$ .

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

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

$c = \sqrt{36 + 0}$

Add $36$ and $0$ .

$c = \sqrt{36}$

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

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

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

$c = 6$

Using the equation $c^{2} = a^{2} + b^{2}$ . Substitute $1$ for $a$ and $6$ for $c$ .

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Rewrite the equation as ${(1)}^{2} + b^{2} = 6^{2}$ .

${(1)}^{2} + b^{2} = 6^{2}$

Simplify each term.

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Remove parentheses around $1$ .

$1^{2} + b^{2} = 6^{2}$

One to any power is one.

$1 + b^{2} = 6^{2}$

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

$1 + b^{2} = 36$

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

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Subtract $1$ from both sides of the equation.

$b^{2} = - 1 + 36$

Add $- 1$ and $36$ .

$b^{2} = 35$

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

$b = \pm \sqrt{35}$

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

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First, use the positive value of the $\pm$ to find the first solution.

$b = \sqrt{35}$

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

$b = - \sqrt{35}$

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

$b = \sqrt{35}, - \sqrt{35}$

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

$b = \sqrt{35}$

The slope of the line between the focus $(- 4, 3)$ and the center $(2, 3)$ determines whether the hyperbola 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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Slope is equal to the change in $y$ over the change in $x$ , or rise over run.

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

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).

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

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

$m = \frac{3 - (3)}{2 - (- 4)}$

Simplify.

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Simplify the numerator.

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Multiply $- 1$ by $3$ .

$m = \frac{3 - 3}{2 - (- 4)}$

Subtract $3$ from $3$ .

$m = \frac{0}{2 - (- 4)}$

Simplify the denominator.

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Remove parentheses.

$m = \frac{0}{2 - (- 4)}$

Multiply $- 1$ by $- 4$ .

$m = \frac{0}{2 + 4}$

Add $2$ and $4$ .

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

Divide $0$ by $6$ .

$m = 0$

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

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

Substitute the values $h = 2$ , $k = 3$ , $a = 1$ , and $b = \sqrt{35}$ into $\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the hyperbola equation $\frac{{(x - (2))}^{2}}{{(1)}^{2}} - \frac{{(y - (3))}^{2}}{{(\sqrt{35})}^{2}} = 1$ .

$\frac{{(x - (2))}^{2}}{{(1)}^{2}} - \frac{{(y - (3))}^{2}}{{(\sqrt{35})}^{2}} = 1$

Simplify to find the final equation of the hyperbola.

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Multiply $- 1$ by $2$ .

$\frac{{(x - 2)}^{2}}{{(1)}^{2}} - \frac{{(y - (3))}^{2}}{{(\sqrt{35})}^{2}} = 1$

Simplify the denominator.

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Remove parentheses around $1$ .

$\frac{{(x - 2)}^{2}}{1^{2}} - \frac{{(y - (3))}^{2}}{{(\sqrt{35})}^{2}} = 1$

One to any power is one.

$\frac{{(x - 2)}^{2}}{1} - \frac{{(y - (3))}^{2}}{{(\sqrt{35})}^{2}} = 1$

Divide ${(x - 2)}^{2}$ by $1$ .

${(x - 2)}^{2} - \frac{{(y - (3))}^{2}}{{(\sqrt{35})}^{2}} = 1$

Multiply $- 1$ by $3$ .

${(x - 2)}^{2} - \frac{{(y - 3)}^{2}}{{(\sqrt{35})}^{2}} = 1$

Simplify the denominator.

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Remove parentheses around $\sqrt{35}$ .

${(x - 2)}^{2} - \frac{{(y - 3)}^{2}}{{\sqrt{35}}^{2}} = 1$

Rewrite ${\sqrt{35}}^{2}$ as $35$ .

${(x - 2)}^{2} - \frac{{(y - 3)}^{2}}{35} = 1$

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