Find Hyperbola: Center (0,0), Focus (0,10), Vertex (0,-6) (0,0) , (0,10) , (0,-6)

$(0, 0)$ , $(0, 10)$ , $(0, - 6)$

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 $(0, - 6)$ and the center point $(0, 0)$ .

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

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

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

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

Simplify.

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Subtract $0$ from $0$ .

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

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

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

Subtract $0$ from $- 6$ .

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

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

$a = \sqrt{0 + 36}$

Add $0$ and $36$ .

$a = \sqrt{36}$

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

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

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

$a = 6$

$c$ is the distance between the focus $(0, 10)$ and the center $(0, 0)$ .

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

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

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

$c = \sqrt{{(0 - 0)}^{2} + {(10 - 0)}^{2}}$

Simplify.

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Subtract $0$ from $0$ .

$c = \sqrt{0^{2} + {(10 - 0)}^{2}}$

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

$c = \sqrt{0 + {(10 - 0)}^{2}}$

Subtract $0$ from $10$ .

$c = \sqrt{0 + 10^{2}}$

Raise $10$ to the power of $2$ .

$c = \sqrt{0 + 100}$

Add $0$ and $100$ .

$c = \sqrt{100}$

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 $6$ for $a$ and $10$ for $c$ .

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Rewrite the equation as ${(6)}^{2} + b^{2} = 10^{2}$ .

${(6)}^{2} + b^{2} = 10^{2}$

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

$36 + b^{2} = 10^{2}$

Raise $10$ to the power of $2$ .

$36 + b^{2} = 100$

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

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Subtract $36$ from both sides of the equation.

$b^{2} = 100 - 36$

Subtract $36$ from $100$ .

$b^{2} = 64$

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

$b = \pm \sqrt{64}$

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

$b = \pm \sqrt{8^{2}}$

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

$b = \pm 8$

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.

$b = 8$

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

$b = - 8$

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

$b = 8, - 8$

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

$b = 8$

The slope of the line between the focus $(0, 10)$ and the center $(0, 0)$ 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.

$m = \frac{0 - (10)}{0 - (0)}$

Simplify.

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Subtract $0$ from $0$ .

$\frac{0 - (10)}{0}$

The expression contains a division by $0$ The expression is undefined.

Undefined

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

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

Substitute the values $h = 0$ , $k = 0$ , $a = 6$ , and $b = 8$ into $\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$ to get the hyperbola equation $\frac{{(y - (0))}^{2}}{{(6)}^{2}} - \frac{{(x - (0))}^{2}}{{(8)}^{2}} = 1$ .

$\frac{{(y - (0))}^{2}}{{(6)}^{2}} - \frac{{(x - (0))}^{2}}{{(8)}^{2}} = 1$

Simplify to find the final equation of the hyperbola.

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Simplify the numerator.

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Multiply $- 1$ by $0$ .

$\frac{{(y + 0)}^{2}}{{(6)}^{2}} - \frac{{(x - (0))}^{2}}{{(8)}^{2}} = 1$

Add $y$ and $0$ .

$\frac{y^{2}}{{(6)}^{2}} - \frac{{(x - (0))}^{2}}{{(8)}^{2}} = 1$

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

$\frac{y^{2}}{36} - \frac{{(x - (0))}^{2}}{{(8)}^{2}} = 1$

Simplify the numerator.

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Multiply $- 1$ by $0$ .

$\frac{y^{2}}{36} - \frac{{(x + 0)}^{2}}{{(8)}^{2}} = 1$

Add $x$ and $0$ .

$\frac{y^{2}}{36} - \frac{x^{2}}{{(8)}^{2}} = 1$

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

$\frac{y^{2}}{36} - \frac{x^{2}}{64} = 1$

Find Hyperbola: Center (0,0), Focus (0,10), Vertex (0,-6) (0,0) , (0,10) , (0,-6)

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