Find the Foci (x^2)/4-(y^2)/40=1

$\frac{x^{2}}{4} - \frac{y^{2}}{40} = 1$

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 $1$ .

$\frac{x^{2}}{4} - \frac{y^{2}}{40} = 1$

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, $a$ .

$a = 2$

$b = 2 \sqrt{10}$

$k = 0$

$h = 0$

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.

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

Simplify.

Simplify the expression.

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

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

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$\sqrt{4 + 4 \cdot 10^{1}}$

Evaluate the exponent.

$\sqrt{4 + 4 \cdot 10}$

Simplify the expression.

Multiply $4$ by $10$ .

$\sqrt{4 + 40}$

Add $4$ and $40$ .

$\sqrt{44}$

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$\sqrt{4 (11)}$

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

$\sqrt{2^{2} \cdot 11}$

Pull terms out from under the radical.

$2 \sqrt{11}$

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.

$(2 \sqrt{11}, 0)$

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.

$(- 2 \sqrt{11}, 0)$

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

$(2 \sqrt{11}, 0), (- 2 \sqrt{11}, 0)$

Step 6

Find the Foci (x^2)/4-(y^2)/40=1

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