Graph x^2+49y^2=49

$x^{2} + 49 y^{2} = 49$

Find the standard form of the ellipse.

Divide each term by $49$ to make the right side equal to one.

$\frac{x^{2}}{49} + \frac{49 y^{2}}{49} = \frac{49}{49}$

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 $1$ .

$\frac{x^{2}}{49} + y^{2} = 1$

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.

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

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.

$a = 7$

$b = 1$

$k = 0$

$h = 0$

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.

$\sqrt{{(7)}^{2} - {(1)}^{2}}$

Simplify.

Raise $7$ to the power of $2$ .

$\sqrt{49 - {(1)}^{2}}$

One to any power is one.

$\sqrt{49 - 1 \cdot 1}$

Multiply $- 1$ by $1$ .

$\sqrt{49 - 1}$

Subtract $1$ from $49$ .

$\sqrt{48}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$\sqrt{16 (3)}$

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2} \cdot 3}$

Pull terms out from under the radical.

$4 \sqrt{3}$

Find the vertices.

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

$(h + a, k)$

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

$(0 + 7, 0)$

Simplify.

$(7, 0)$

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

$(h - a, k)$

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

$(0 - (7), 0)$

Simplify.

$(- 7, 0)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(7, 0)$

${Vertex}_{2}$ : $(- 7, 0)$

${Vertex}_{1}$ : $(7, 0)$

${Vertex}_{2}$ : $(- 7, 0)$

Find the foci.

The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

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

$(0 + 4 \sqrt{3}, 0)$

Simplify.

$(4 \sqrt{3}, 0)$

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

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

$(0 - (4 \sqrt{3}), 0)$

Simplify.

$(- 4 \sqrt{3}, 0)$

Ellipses have two foci.

${Focus}_{1}$ : $(4 \sqrt{3}, 0)$

${Focus}_{2}$ : $(- 4 \sqrt{3}, 0)$

${Focus}_{1}$ : $(4 \sqrt{3}, 0)$

${Focus}_{2}$ : $(- 4 \sqrt{3}, 0)$

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.

$\frac{\sqrt{{(7)}^{2} - {(1)}^{2}}}{7}$

Simplify the numerator.

Raise $7$ to the power of $2$ .

$\frac{\sqrt{49 - 1^{2}}}{7}$

One to any power is one.

$\frac{\sqrt{49 - 1 \cdot 1}}{7}$

Multiply $- 1$ by $1$ .

$\frac{\sqrt{49 - 1}}{7}$

Subtract $1$ from $49$ .

$\frac{\sqrt{48}}{7}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$\frac{\sqrt{16 (3)}}{7}$

Rewrite $16$ as $4^{2}$ .

$\frac{\sqrt{4^{2} \cdot 3}}{7}$

Pull terms out from under the radical.

$\frac{4 \sqrt{3}}{7}$

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

Center: $(0, 0)$

${Vertex}_{1}$ : $(7, 0)$

${Vertex}_{2}$ : $(- 7, 0)$

${Focus}_{1}$ : $(4 \sqrt{3}, 0)$

${Focus}_{2}$ : $(- 4 \sqrt{3}, 0)$

Eccentricity: $\frac{4 \sqrt{3}}{7}$

Graph x^2+49y^2=49

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