Find the Foci ((x+5)^2)/36-((y+1)^2)/64=1

$\frac{{(x + 5)}^{2}}{36} - \frac{{(y + 1)}^{2}}{64} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x + 5)}^{2}}{36} - \frac{{(y + 1)}^{2}}{64} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

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

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 6$

$b = 8$

$k = - 1$

$h = - 5$

Step 4

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(6)}^{2} + {(8)}^{2}}$

Simplify.

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

$\sqrt{36 + {(8)}^{2}}$

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

$\sqrt{36 + 64}$

Add $36$ and $64$ .

$\sqrt{100}$

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

$\sqrt{10^{2}}$

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

$10$

Step 5

Find the foci.

The first focus of a hyperbola can be found by adding $c$ to $h$ .

$(h + c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(5, - 1)$

The second focus of a hyperbola can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(- 15, - 1)$

The foci of a hyperbola follow the form of $(h \pm \sqrt{a^{2} + b^{2}}, k)$ . Hyperbolas have two foci.

$(5, - 1), (- 15, - 1)$

Step 6

Find the Foci ((x+5)^2)/36-((y+1)^2)/64=1

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