Find the Properties 25x^2+y^2=400

$25 x^{2} + y^{2} = 400$

Find the standard form of the ellipse.

Divide each term by $400$ to make the right side equal to one.

$\frac{25 x^{2}}{400} + \frac{y^{2}}{400} = \frac{400}{400}$

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 $1$ .

$\frac{x^{2}}{16} + \frac{y^{2}}{400} = 1$

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.

$a = 20$

$b = 4$

$k = 0$

$h = 0$

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

$(0, 0)$

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.

$\sqrt{{(20)}^{2} - {(4)}^{2}}$

Simplify.

Raise $20$ to the power of $2$ .

$\sqrt{400 - {(4)}^{2}}$

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

$\sqrt{400 - 1 \cdot 16}$

Multiply $- 1$ by $16$ .

$\sqrt{400 - 16}$

Subtract $16$ from $400$ .

$\sqrt{384}$

Rewrite $384$ as $8^{2} \cdot 6$ .

Factor $64$ out of $384$ .

$\sqrt{64 (6)}$

Rewrite $64$ as $8^{2}$ .

$\sqrt{8^{2} \cdot 6}$

Pull terms out from under the radical.

$8 \sqrt{6}$

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.

$(0, 0 + 20)$

Simplify.

$(0, 20)$

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.

$(0, 0 - (20))$

Simplify.

$(0, - 20)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, 20)$

${Vertex}_{2}$ : $(0, - 20)$

${Vertex}_{1}$ : $(0, 20)$

${Vertex}_{2}$ : $(0, - 20)$

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.

$(0, 0 + 8 \sqrt{6})$

Simplify.

$(0, 8 \sqrt{6})$

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.

$(0, 0 - (8 \sqrt{6}))$

Simplify.

$(0, - 8 \sqrt{6})$

Ellipses have two foci.

${Focus}_{1}$ : $(0, 8 \sqrt{6})$

${Focus}_{2}$ : $(0, - 8 \sqrt{6})$

${Focus}_{1}$ : $(0, 8 \sqrt{6})$

${Focus}_{2}$ : $(0, - 8 \sqrt{6})$

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.

$\frac{\sqrt{{(20)}^{2} - {(4)}^{2}}}{20}$

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$\frac{\sqrt{400 - 4^{2}}}{20}$

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

$\frac{\sqrt{400 - 1 \cdot 16}}{20}$

Multiply $- 1$ by $16$ .

$\frac{\sqrt{400 - 16}}{20}$

Subtract $16$ from $400$ .

$\frac{\sqrt{384}}{20}$

Rewrite $384$ as $8^{2} \cdot 6$ .

Factor $64$ out of $384$ .

$\frac{\sqrt{64 (6)}}{20}$

Rewrite $64$ as $8^{2}$ .

$\frac{\sqrt{8^{2} \cdot 6}}{20}$

Pull terms out from under the radical.

$\frac{8 \sqrt{6}}{20}$

Cancel the common factor of $8$ and $20$ .

Factor $4$ out of $8 \sqrt{6}$ .

$\frac{4 (2 \sqrt{6})}{20}$

Cancel the common factors.

Factor $4$ out of $20$ .

$\frac{4 (2 \sqrt{6})}{4 (5)}$

Cancel the common factor.

$\frac{4 (2 \sqrt{6})}{4 \cdot 5}$

Rewrite the expression.

$\frac{2 \sqrt{6}}{5}$

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

Center: $(0, 0)$

${Vertex}_{1}$ : $(0, 20)$

${Vertex}_{2}$ : $(0, - 20)$

${Focus}_{1}$ : $(0, 8 \sqrt{6})$

${Focus}_{2}$ : $(0, - 8 \sqrt{6})$

Eccentricity: $\frac{2 \sqrt{6}}{5}$

Find the Properties 25x^2+y^2=400

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