Find the Foci ((x-3)^2)/9+((y-6)^2)/16=1

$\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 6)}^{2}}{16} = 1$

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 $1$ .

$\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 6)}^{2}}{16} = 1$

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.

$a = 4$

$b = 3$

$k = 6$

$h = 3$

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.

$\sqrt{{(4)}^{2} - {(3)}^{2}}$

Simplify.

Raise $4$ to the power of $2$ .

$\sqrt{16 - {(3)}^{2}}$

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

$\sqrt{16 - 1 \cdot 9}$

Multiply $- 1$ by $9$ .

$\sqrt{16 - 9}$

Subtract $9$ from $16$ .

$\sqrt{7}$

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.

$(3, 6 + \sqrt{7})$

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.

$(3, 6 - (\sqrt{7}))$

Simplify.

$(3, 6 - \sqrt{7})$

Ellipses have two foci.

${Focus}_{1}$ : $(3, 6 + \sqrt{7})$

${Focus}_{2}$ : $(3, 6 - \sqrt{7})$

${Focus}_{1}$ : $(3, 6 + \sqrt{7})$

${Focus}_{2}$ : $(3, 6 - \sqrt{7})$

Step 6

Find the Foci ((x-3)^2)/9+((y-6)^2)/16=1

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