Find the Foci ((y-2)^2)/16-((x+1)^2)/144=1

$\frac{{(y - 2)}^{2}}{16} - \frac{{(x + 1)}^{2}}{144} = 1$

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 $1$ .

$\frac{{(y - 2)}^{2}}{16} - \frac{{(x + 1)}^{2}}{144} = 1$

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.

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

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, $a$ .

$a = 4$

$b = 12$

$k = 2$

$h = - 1$

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.

$\sqrt{{(4)}^{2} + {(12)}^{2}}$

Simplify.

Raise $4$ to the power of $2$ .

$\sqrt{16 + {(12)}^{2}}$

Raise $12$ to the power of $2$ .

$\sqrt{16 + 144}$

Add $16$ and $144$ .

$\sqrt{160}$

Rewrite $160$ as $4^{2} \cdot 10$ .

Factor $16$ out of $160$ .

$\sqrt{16 (10)}$

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2} \cdot 10}$

Pull terms out from under the radical.

$4 \sqrt{10}$

Step 5

Find the foci.

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

$(h, k + c)$

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

$(- 1, 2 + 4 \sqrt{10})$

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

$(h, k - c)$

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

$(- 1, 2 - 4 \sqrt{10})$

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

$(- 1, 2 + 4 \sqrt{10}), (- 1, 2 - 4 \sqrt{10})$

Step 6

Find the Foci ((y-2)^2)/16-((x+1)^2)/144=1

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