Algebra Examples

$(5, 1)$ , $(4, 1)$ , $(- 5, 1)$

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 $(4, 1)$ and the center point $(5, 1)$ .

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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.

$a = \sqrt{{((4) - (5))}^{2} + {((1) - (1))}^{2}}$

Simplify.

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Multiply $- 1$ by $5$ to get $- 5$ .

$a = \sqrt{{(4 - 5)}^{2} + {((1) - (1))}^{2}}$

Subtract $5$ from $4$ to get $- 1$ .

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

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

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

Multiply $- 1$ by $1$ to get $- 1$ .

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

Subtract $1$ from $1$ to get $0$ .

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

Remove parentheses around $0$ .

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

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

$a = \sqrt{1 + 0}$

Add $1$ and $0$ to get $1$ .

$a = \sqrt{1}$

Any root of $1$ is $1$ .

$a = 1$

$c$ is the distance between the focus $(- 5, 1)$ and the center $(5, 1)$ .

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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.

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

Simplify.

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Multiply $- 1$ by $5$ to get $- 5$ .

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

Subtract $5$ from $- 5$ to get $- 10$ .

$c = \sqrt{{(- 10)}^{2} + {((1) - (1))}^{2}}$

Raise $- 10$ to the power of $2$ to get $100$ .

$c = \sqrt{100 + {((1) - (1))}^{2}}$

Multiply $- 1$ by $1$ to get $- 1$ .

$c = \sqrt{100 + {(1 - 1)}^{2}}$

Subtract $1$ from $1$ to get $0$ .

$c = \sqrt{100 + {(0)}^{2}}$

Remove parentheses around $0$ .

$c = \sqrt{100 + 0^{2}}$

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

$c = \sqrt{100 + 0}$

Add $100$ and $0$ to get $100$ .

$c = \sqrt{100}$

Rewrite $100$ as $10^{2}$ .

$c = \sqrt{10^{2}}$

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

$c = 10$

Using the equation $c^{2} = a^{2} + b^{2}$ . Substitute $1$ for $a$ and $10$ for $c$ to get ${(10)}^{2} = {(1)}^{2} + b^{2}$ .

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Move $b$ to the right side of the equation by subtracting $b$ from both sides of the equation.

$b^{2} = 99$

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

$b = \pm \sqrt{99}$

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

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Simplify the right side of the equation.

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Rewrite $99$ as $3^{2} \cdot 11$ .

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Factor $9$ out of $99$ .

$b = \pm \sqrt{9 (11)}$

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

$b = \pm \sqrt{3^{2} \cdot 11}$

Pull terms out from under the radical.

$b = \pm 3 \sqrt{11}$

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.

$b = 3 \sqrt{11}$

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

$b = - 3 \sqrt{11}$

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

$b = 3 \sqrt{11}, - 3 \sqrt{11}$

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

$b = 3 \sqrt{11}$

The slope of the line between the focus $(- 5, 1)$ and the center $(5, 1)$ 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.

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

Simplify.

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Simplify the numerator.

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Multiply $- 1$ by $1$ to get $- 1$ .

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

Subtract $1$ from $1$ to get $0$ .

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

Simplify the denominator.

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Remove parentheses.

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

Multiply $- 1$ by $- 5$ to get $5$ .

$m = \frac{0}{5 + 5}$

Add $5$ and $5$ to get $10$ .

$m = \frac{0}{10}$

Divide $0$ by $10$ to get $0$ .

$m = 0$

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 $h = 5$ , $k = 1$ , $a = 1$ , and $b = 3 \sqrt{11}$ into $\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the hyperbola equation $\frac{{(x - (5))}^{2}}{{(1)}^{2}} - \frac{{(y - (1))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$ .

$\frac{{(x - (5))}^{2}}{{(1)}^{2}} - \frac{{(y - (1))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Simplify to find the final equation of the hyperbola.

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Multiply $- 1$ by $5$ to get $- 5$ .

$\frac{{(x - 5)}^{2}}{{(1)}^{2}} - \frac{{(y - (1))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Simplify the denominator.

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Remove parentheses around $1$ .

$\frac{{(x - 5)}^{2}}{1^{2}} - \frac{{(y - (1))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

One to any power is one.

$\frac{{(x - 5)}^{2}}{1} - \frac{{(y - (1))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Divide ${(x - 5)}^{2}$ by $1$ to get ${(x - 5)}^{2}$ .

${(x - 5)}^{2} - \frac{{(y - (1))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Multiply $- 1$ by $1$ to get $- 1$ .

${(x - 5)}^{2} - \frac{{(y - 1)}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Simplify the denominator.

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Apply the product rule to $3 \sqrt{11}$ .

${(x - 5)}^{2} - \frac{{(y - 1)}^{2}}{3^{2} {\sqrt{11}}^{2}} = 1$

Raise $3$ to the power of $2$ to get $9$ .

${(x - 5)}^{2} - \frac{{(y - 1)}^{2}}{9 {\sqrt{11}}^{2}} = 1$

Rewrite ${\sqrt{11}}^{2}$ as $11$ .

${(x - 5)}^{2} - \frac{{(y - 1)}^{2}}{9 \cdot 11} = 1$

Multiply $9$ by $11$ to get $99$ .

${(x - 5)}^{2} - \frac{{(y - 1)}^{2}}{99} = 1$

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