Find the Foci 9y^2+x^2=36

$9 y^{2} + x^{2} = 36$

Step 1

Find the standard form of the ellipse.

Divide each term by $36$ to make the right side equal to one.

$\frac{9 y^{2}}{36} + \frac{x^{2}}{36} = \frac{36}{36}$

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 $1$ .

$\frac{y^{2}}{4} + \frac{x^{2}}{36} = 1$

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.

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

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.

$a = 6$

$b = 2$

$k = 0$

$h = 0$

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.

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

Simplify.

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

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

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

$\sqrt{36 - 1 \cdot 4}$

Multiply $- 1$ by $4$ .

$\sqrt{36 - 4}$

Subtract $4$ from $36$ .

$\sqrt{32}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$\sqrt{16 (2)}$

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2} \cdot 2}$

Pull terms out from under the radical.

$4 \sqrt{2}$

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.

$(0 + 4 \sqrt{2}, 0)$

Simplify.

$(4 \sqrt{2}, 0)$

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.

$(0 - (4 \sqrt{2}), 0)$

Simplify.

$(- 4 \sqrt{2}, 0)$

Ellipses have two foci.

${Focus}_{1}$ : $(4 \sqrt{2}, 0)$

${Focus}_{2}$ : $(- 4 \sqrt{2}, 0)$

${Focus}_{1}$ : $(4 \sqrt{2}, 0)$

${Focus}_{2}$ : $(- 4 \sqrt{2}, 0)$

Step 6

Find the Foci 9y^2+x^2=36

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