Find the Foci 10x^2-36y^2=100

$10 x^{2} - 36 y^{2} = 100$

Step 1

Find the standard form of the hyperbola.

Divide each term by $100$ to make the right side equal to one.

$\frac{10 x^{2}}{100} - \frac{36 y^{2}}{100} = \frac{100}{100}$

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 $1$ .

$\frac{x^{2}}{10} - \frac{y^{2}}{\frac{25}{9}} = 1$

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, $a$ .

$a = \sqrt{10}$

$b = \frac{5}{3}$

$k = 0$

$h = 0$

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.

$\sqrt{{(\sqrt{10})}^{2} + {(\frac{5}{3})}^{2}}$

Simplify.

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

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

$\sqrt{{(10^{\frac{1}{2}})}^{2} + {(\frac{5}{3})}^{2}}$

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

$\sqrt{10^{\frac{1}{2} \cdot 2} + {(\frac{5}{3})}^{2}}$

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

$\sqrt{10^{\frac{2}{2}} + {(\frac{5}{3})}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{10^{\frac{2}{2}} + {(\frac{5}{3})}^{2}}$

Rewrite the expression.

$\sqrt{10^{1} + {(\frac{5}{3})}^{2}}$

Evaluate the exponent.

$\sqrt{10 + {(\frac{5}{3})}^{2}}$

Simplify the expression.

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

$\sqrt{10 + \frac{5^{2}}{3^{2}}}$

Raise $5$ to the power of $2$ .

$\sqrt{10 + \frac{25}{3^{2}}}$

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

$\sqrt{10 + \frac{25}{9}}$

To write $10$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\sqrt{10 \cdot \frac{9}{9} + \frac{25}{9}}$

Combine $10$ and $\frac{9}{9}$ .

$\sqrt{\frac{10 \cdot 9}{9} + \frac{25}{9}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{10 \cdot 9 + 25}{9}}$

Simplify the numerator.

Multiply $10$ by $9$ .

$\sqrt{\frac{90 + 25}{9}}$

Add $90$ and $25$ .

$\sqrt{\frac{115}{9}}$

Rewrite $\sqrt{\frac{115}{9}}$ as $\frac{\sqrt{115}}{\sqrt{9}}$ .

$\frac{\sqrt{115}}{\sqrt{9}}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$\frac{\sqrt{115}}{\sqrt{3^{2}}}$

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

$\frac{\sqrt{115}}{3}$

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.

$(\frac{\sqrt{115}}{3}, 0)$

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.

$(- \frac{\sqrt{115}}{3}, 0)$

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

$(\frac{\sqrt{115}}{3}, 0), (- \frac{\sqrt{115}}{3}, 0)$

Step 6

Find the Foci 10x^2-36y^2=100

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