Find the Foci 36x^2+5y^2=180

$36 x^{2} + 5 y^{2} = 180$

Step 1

Find the standard form of the ellipse.

Divide each term by $180$ to make the right side equal to one.

$\frac{36 x^{2}}{180} + \frac{5 y^{2}}{180} = \frac{180}{180}$

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 $1$ .

$\frac{x^{2}}{5} + \frac{y^{2}}{36} = 1$

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.

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

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.

$a = 6$

$b = \sqrt{5}$

$k = 0$

$h = 0$

Step 4

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

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

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

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

$\sqrt{{(6)}^{2} - {(\sqrt{5})}^{2}}$

Simplify.

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

$\sqrt{36 - {(\sqrt{5})}^{2}}$

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

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

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

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

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

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

$\sqrt{36 - 5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{36 - 5^{\frac{2}{2}}}$

Rewrite the expression.

$\sqrt{36 - 5^{1}}$

Evaluate the exponent.

$\sqrt{36 - 1 \cdot 5}$

Simplify the expression.

Multiply $- 1$ by $5$ .

$\sqrt{36 - 5}$

Subtract $5$ from $36$ .

$\sqrt{31}$

Step 5

Find the foci.

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

$(h, k + c)$

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

$(0, 0 + \sqrt{31})$

Simplify.

$(0, \sqrt{31})$

The first focus of an ellipse can be found by subtracting $c$ from $k$ .

$(h, k - c)$

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

$(0, 0 - (\sqrt{31}))$

Simplify.

$(0, - \sqrt{31})$

Ellipses have two foci.

${Focus}_{1}$ : $(0, \sqrt{31})$

${Focus}_{2}$ : $(0, - \sqrt{31})$

${Focus}_{1}$ : $(0, \sqrt{31})$

${Focus}_{2}$ : $(0, - \sqrt{31})$

Step 6

Find the Foci 36x^2+5y^2=180

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