Find the Foci 16y^2-9x^2=144

$16 y^{2} - 9 x^{2} = 144$

Find the standard form of the hyperbola.

Divide each term by $144$ to make the right side equal to one.

$\frac{16 y^{2}}{144} - \frac{9 x^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$

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

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

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, $a$ .

$a = 3$

$b = 4$

$k = 0$

$h = 0$

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.

$\sqrt{{(3)}^{2} + {(4)}^{2}}$

Simplify.

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

$\sqrt{9 + {(4)}^{2}}$

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

$\sqrt{9 + 16}$

Add $9$ and $16$ .

$\sqrt{25}$

Rewrite $25$ as $5^{2}$ .

$\sqrt{5^{2}}$

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

$5$

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.

$(0, 5)$

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.

$(0, - 5)$

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

$(0, 5), (0, - 5)$

Find the Foci 16y^2-9x^2=144

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