Find the Hyperbola: Center (5,6), Focus (-5,6), Vertex (4,6) (5,6) , (4,6) , (-5,6)

$(5, 6)$ , $(4, 6)$ , $(- 5, 6)$

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

Step 2

$a$ is the distance between the vertex $(4, 6)$ and the center point $(5, 6)$ .

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.

$a = \sqrt{{(4 - 5)}^{2} + {(6 - 6)}^{2}}$

Simplify.

Subtract $5$ from $4$ .

$a = \sqrt{{(- 1)}^{2} + {(6 - 6)}^{2}}$

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

$a = \sqrt{1 + {(6 - 6)}^{2}}$

Subtract $6$ from $6$ .

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

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

$a = \sqrt{1 + 0}$

Add $1$ and $0$ .

$a = \sqrt{1}$

Any root of $1$ is $1$ .

$a = 1$

Step 3

$c$ is the distance between the focus $(- 5, 6)$ and the center $(5, 6)$ .

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.

$c = \sqrt{{((- 5) - 5)}^{2} + {(6 - 6)}^{2}}$

Simplify.

Subtract $5$ from $- 5$ .

$c = \sqrt{{(- 10)}^{2} + {(6 - 6)}^{2}}$

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

$c = \sqrt{100 + {(6 - 6)}^{2}}$

Subtract $6$ from $6$ .

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

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

$c = \sqrt{100 + 0}$

Add $100$ and $0$ .

$c = \sqrt{100}$

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

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

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

$c = 10$

Step 4

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

Rewrite the equation as ${(1)}^{2} + b^{2} = 10^{2}$ .

${(1)}^{2} + b^{2} = 10^{2}$

One to any power is one.

$1 + b^{2} = 10^{2}$

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

$1 + b^{2} = 100$

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

Subtract $1$ from both sides of the equation.

$b^{2} = 100 - 1$

Subtract $1$ from $100$ .

$b^{2} = 99$

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

$b = \pm \sqrt{99}$

Simplify $\pm \sqrt{99}$ .

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

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.

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.

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

Step 5

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

$b = 3 \sqrt{11}$

Step 6

The slope of the line between the focus $(- 5, 6)$ and the center $(5, 6)$ 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.

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.

$m = \frac{6 - (6)}{5 - (- 5)}$

Simplify.

Simplify the numerator.

Multiply $- 1$ by $6$ .

$m = \frac{6 - 6}{5 - (- 5)}$

Subtract $6$ from $6$ .

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

Simplify the denominator.

Multiply $- 1$ by $- 5$ .

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

Add $5$ and $5$ .

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

Divide $0$ by $10$ .

$m = 0$

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

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

Step 7

Substitute the values $h = 5$ , $k = 6$ , $a = 1$ , and $b = 3 \sqrt{11}$ into $\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the hyperbola equation $\frac{{(x - (5))}^{2}}{{(1)}^{2}} - \frac{{(y - (6))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$ .

$\frac{{(x - (5))}^{2}}{{(1)}^{2}} - \frac{{(y - (6))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Step 8

Simplify to find the final equation of the hyperbola.

Multiply $- 1$ by $5$ .

$\frac{{(x - 5)}^{2}}{1^{2}} - \frac{{(y - (6))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

One to any power is one.

$\frac{{(x - 5)}^{2}}{1} - \frac{{(y - (6))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

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

${(x - 5)}^{2} - \frac{{(y - (6))}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Multiply $- 1$ by $6$ .

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{{(3 \sqrt{11})}^{2}} = 1$

Simplify the denominator.

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

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{3^{2} {\sqrt{11}}^{2}} = 1$

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

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 {\sqrt{11}}^{2}} = 1$

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{11}$ as $11^{\frac{1}{2}}$ .

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 {(11^{\frac{1}{2}})}^{2}} = 1$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 \cdot 11^{\frac{1}{2} \cdot 2}} = 1$

Combine $\frac{1}{2}$ and $2$ .

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 \cdot 11^{\frac{2}{2}}} = 1$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 \cdot 11^{\frac{2}{2}}} = 1$

Rewrite the expression.

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 \cdot 11} = 1$

Evaluate the exponent.

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{9 \cdot 11} = 1$

Multiply $9$ by $11$ .

${(x - 5)}^{2} - \frac{{(y - 6)}^{2}}{99} = 1$

Step 9

Find the Hyperbola: Center (5,6), Focus (-5,6), Vertex (4,6) (5,6) , (4,6) , (-5,6)

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