Find the Foci 36x^2+9y^2=324

$36 x^{2} + 9 y^{2} = 324$

Step 1

Find the standard form of the ellipse.

Divide each term by $324$ to make the right side equal to one.

$\frac{36 x^{2}}{324} + \frac{9 y^{2}}{324} = \frac{324}{324}$

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 $1$ .

$\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$

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.

$a = 6$

$b = 3$

$k = 0$

$h = 0$

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.

$\sqrt{{(6)}^{2} - {(3)}^{2}}$

Simplify.

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

$\sqrt{36 - {(3)}^{2}}$

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

$\sqrt{36 - 1 \cdot 9}$

Multiply $- 1$ by $9$ .

$\sqrt{36 - 9}$

Subtract $9$ from $36$ .

$\sqrt{27}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$\sqrt{9 (3)}$

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

$\sqrt{3^{2} \cdot 3}$

Pull terms out from under the radical.

$3 \sqrt{3}$

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.

$(0, 0 + 3 \sqrt{3})$

Simplify.

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

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.

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

Simplify.

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

Ellipses have two foci.

${Focus}_{1}$ : $(0, 3 \sqrt{3})$

${Focus}_{2}$ : $(0, - 3 \sqrt{3})$

${Focus}_{1}$ : $(0, 3 \sqrt{3})$

${Focus}_{2}$ : $(0, - 3 \sqrt{3})$

Step 6

Find the Foci 36x^2+9y^2=324

Related Questions

Support LCMGCF.com by disabling your adblocker.