Find the Foci (x^2)/109+(y^2)/9=1

$\frac{x^{2}}{109} + \frac{y^{2}}{9} = 1$

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 $1$ .

$\frac{x^{2}}{109} + \frac{y^{2}}{9} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

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

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = \sqrt{109}$

$b = 3$

$k = 0$

$h = 0$

Step 4

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the ellipse by using the following formula.

$\sqrt{a^{2} - b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(\sqrt{109})}^{2} - {(3)}^{2}}$

Simplify.

Rewrite ${\sqrt{109}}^{2}$ as $109$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{109}$ as $109^{\frac{1}{2}}$ .

$\sqrt{{(109^{\frac{1}{2}})}^{2} - {(3)}^{2}}$

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

$\sqrt{109^{\frac{1}{2} \cdot 2} - {(3)}^{2}}$

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

$\sqrt{109^{\frac{2}{2}} - {(3)}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{109^{\frac{2}{2}} - {(3)}^{2}}$

Rewrite the expression.

$\sqrt{109^{1} - {(3)}^{2}}$

Evaluate the exponent.

$\sqrt{109 - {(3)}^{2}}$

Simplify the expression.

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

$\sqrt{109 - 1 \cdot 9}$

Multiply $- 1$ by $9$ .

$\sqrt{109 - 9}$

Subtract $9$ from $109$ .

$\sqrt{100}$

Rewrite $100$ as $10^{2}$ .

$\sqrt{10^{2}}$

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

$10$

Step 5

Find the foci.

The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(0 + 10, 0)$

Simplify.

$(10, 0)$

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(0 - (10), 0)$

Simplify.

$(- 10, 0)$

Ellipses have two foci.

${Focus}_{1}$ : $(10, 0)$

${Focus}_{2}$ : $(- 10, 0)$

${Focus}_{1}$ : $(10, 0)$

${Focus}_{2}$ : $(- 10, 0)$

Step 6

Find the Foci (x^2)/109+(y^2)/9=1

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