Find the Foci (x^2)/40-(y^2)/81=1

$\frac{x^{2}}{40} - \frac{y^{2}}{81} = 1$

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 $1$ .

$\frac{x^{2}}{40} - \frac{y^{2}}{81} = 1$

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, $a$ .

$a = 2 \sqrt{10}$

$b = 9$

$k = 0$

$h = 0$

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.

$\sqrt{{(2 \sqrt{10})}^{2} + {(9)}^{2}}$

Simplify.

Simplify the expression.

Apply the product rule to $2 \sqrt{10}$ .

$\sqrt{2^{2} {\sqrt{10}}^{2} + {(9)}^{2}}$

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

$\sqrt{4 {\sqrt{10}}^{2} + {(9)}^{2}}$

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

$\sqrt{4 {(10^{\frac{1}{2}})}^{2} + {(9)}^{2}}$

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

$\sqrt{4 \cdot 10^{\frac{1}{2} \cdot 2} + {(9)}^{2}}$

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

$\sqrt{4 \cdot 10^{\frac{2}{2}} + {(9)}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{4 \cdot 10^{\frac{2}{2}} + {(9)}^{2}}$

Rewrite the expression.

$\sqrt{4 \cdot 10^{1} + {(9)}^{2}}$

Evaluate the exponent.

$\sqrt{4 \cdot 10 + {(9)}^{2}}$

Simplify the expression.

Multiply $4$ by $10$ .

$\sqrt{40 + {(9)}^{2}}$

Raise $9$ to the power of $2$ .

$\sqrt{40 + 81}$

Add $40$ and $81$ .

$\sqrt{121}$

Rewrite $121$ as $11^{2}$ .

$\sqrt{11^{2}}$

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

$11$

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.

$(11, 0)$

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.

$(- 11, 0)$

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

$(11, 0), (- 11, 0)$

Step 6

Find the Foci (x^2)/40-(y^2)/81=1

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