Find the Foci ((x-7)^2)/4+((y+3)^2)/16=1

$\frac{{(x - 7)}^{2}}{4} + \frac{{(y + 3)}^{2}}{16} = 1$

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 $1$ .

$\frac{{(x - 7)}^{2}}{4} + \frac{{(y + 3)}^{2}}{16} = 1$

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.

$a = 4$

$b = 2$

$k = - 3$

$h = 7$

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.

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

Simplify.

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

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

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

$\sqrt{16 - 1 \cdot 4}$

Multiply $- 1$ by $4$ .

$\sqrt{16 - 4}$

Subtract $4$ from $16$ .

$\sqrt{12}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$\sqrt{4 (3)}$

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

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

Pull terms out from under the radical.

$2 \sqrt{3}$

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.

$(7, - 3 + 2 \sqrt{3})$

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.

$(7, - 3 - (2 \sqrt{3}))$

Simplify.

$(7, - 3 - 2 \sqrt{3})$

Ellipses have two foci.

${Focus}_{1}$ : $(7, - 3 + 2 \sqrt{3})$

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

${Focus}_{1}$ : $(7, - 3 + 2 \sqrt{3})$

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

Step 6

Find the Foci ((x-7)^2)/4+((y+3)^2)/16=1

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