Find the Foci (y^2)/36-(x^2)/45=1

$\frac{y^{2}}{36} - \frac{x^{2}}{45} = 1$

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 $1$ .

$\frac{y^{2}}{36} - \frac{x^{2}}{45} = 1$

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, $a$ .

$a = 6$

$b = 3 \sqrt{5}$

$k = 0$

$h = 0$

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.

$\sqrt{{(6)}^{2} + {(3 \sqrt{5})}^{2}}$

Simplify.

Simplify the expression.

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

$\sqrt{36 + {(3 \sqrt{5})}^{2}}$

Apply the product rule to $3 \sqrt{5}$ .

$\sqrt{36 + 3^{2} {\sqrt{5}}^{2}}$

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

$\sqrt{36 + 9 {\sqrt{5}}^{2}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$\sqrt{36 + 9 {(5^{\frac{1}{2}})}^{2}}$

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

$\sqrt{36 + 9 \cdot 5^{\frac{1}{2} \cdot 2}}$

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

$\sqrt{36 + 9 \cdot 5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{36 + 9 \cdot 5^{\frac{2}{2}}}$

Rewrite the expression.

$\sqrt{36 + 9 \cdot 5^{1}}$

Evaluate the exponent.

$\sqrt{36 + 9 \cdot 5}$

Simplify the expression.

Multiply $9$ by $5$ .

$\sqrt{36 + 45}$

Add $36$ and $45$ .

$\sqrt{81}$

Rewrite $81$ as $9^{2}$ .

$\sqrt{9^{2}}$

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

$9$

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.

$(0, 9)$

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.

$(0, - 9)$

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

$(0, 9), (0, - 9)$

Step 6

Find the Foci (y^2)/36-(x^2)/45=1

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