Find the Properties (x^2)/64+((y-6)^2)/121=1

$\frac{x^{2}}{64} + \frac{{(y - 6)}^{2}}{121} = 1$

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 $1$ .

$\frac{x^{2}}{64} + \frac{{(y - 6)}^{2}}{121} = 1$

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.

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

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.

$a = 11$

$b = 8$

$k = 6$

$h = 0$

The center of an ellipse follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(0, 6)$

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.

$\sqrt{{(11)}^{2} - {(8)}^{2}}$

Simplify.

Raise $11$ to the power of $2$ .

$\sqrt{121 - {(8)}^{2}}$

Raise $8$ to the power of $2$ .

$\sqrt{121 - 1 \cdot 64}$

Multiply $- 1$ by $64$ .

$\sqrt{121 - 64}$

Subtract $64$ from $121$ .

$\sqrt{57}$

Find the vertices.

The first vertex of an ellipse can be found by adding $a$ to $k$ .

$(h, k + a)$

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

$(0, 6 + 11)$

Simplify.

$(0, 17)$

The second vertex of an ellipse can be found by subtracting $a$ from $k$ .

$(h, k - a)$

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

$(0, 6 - (11))$

Simplify.

$(0, - 5)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, 17)$

${Vertex}_{2}$ : $(0, - 5)$

${Vertex}_{1}$ : $(0, 17)$

${Vertex}_{2}$ : $(0, - 5)$

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.

$(0, 6 + \sqrt{57})$

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.

$(0, 6 - (\sqrt{57}))$

Simplify.

$(0, 6 - \sqrt{57})$

Ellipses have two foci.

${Focus}_{1}$ : $(0, 6 + \sqrt{57})$

${Focus}_{2}$ : $(0, 6 - \sqrt{57})$

${Focus}_{1}$ : $(0, 6 + \sqrt{57})$

${Focus}_{2}$ : $(0, 6 - \sqrt{57})$

Find the eccentricity.

Find the eccentricity by using the following formula.

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

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

$\frac{\sqrt{{(11)}^{2} - {(8)}^{2}}}{11}$

Simplify the numerator.

Raise $11$ to the power of $2$ .

$\frac{\sqrt{121 - 8^{2}}}{11}$

Raise $8$ to the power of $2$ .

$\frac{\sqrt{121 - 1 \cdot 64}}{11}$

Multiply $- 1$ by $64$ .

$\frac{\sqrt{121 - 64}}{11}$

Subtract $64$ from $121$ .

$\frac{\sqrt{57}}{11}$

These values represent the important values for graphing and analyzing an ellipse.

Center: $(0, 6)$

${Vertex}_{1}$ : $(0, 17)$

${Vertex}_{2}$ : $(0, - 5)$

${Focus}_{1}$ : $(0, 6 + \sqrt{57})$

${Focus}_{2}$ : $(0, 6 - \sqrt{57})$

Eccentricity: $\frac{\sqrt{57}}{11}$

Find the Properties (x^2)/64+((y-6)^2)/121=1

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