Precalculus Examples

$y^{2} = 1 - 25 x^{2}$

Step 1

Find the standard form of the ellipse.

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Step 1.1

Move all terms containing variables to the left side of the equation.

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Step 1.1.1

Add $25 x^{2}$ to both sides of the equation.

$y^{2} + 25 x^{2} = 1$

Step 1.1.2

Reorder $y^{2}$ and $25 x^{2}$ .

$25 x^{2} + y^{2} = 1$

Step 1.2

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 $1$ .

$\frac{x^{2}}{\frac{1}{25}} + y^{2} = 1$

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.

$a = 1$

$b = \frac{1}{5}$

$k = 0$

$h = 0$

Step 4

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

$(0, 0)$

Step 5

Find $c$ , the distance from the center to a focus.

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Step 5.1

Find the distance from the center to a focus of the ellipse by using the following formula.

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

Step 5.2

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

$\sqrt{{(1)}^{2} - {(\frac{1}{5})}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

One to any power is one.

$\sqrt{1 - {(\frac{1}{5})}^{2}}$

Step 5.3.2

Apply the product rule to $\frac{1}{5}$ .

$\sqrt{1 - \frac{1^{2}}{5^{2}}}$

Step 5.3.3

One to any power is one.

$\sqrt{1 - \frac{1}{5^{2}}}$

Step 5.3.4

Raise $5$ to the power of $2$ .

$\sqrt{1 - \frac{1}{25}}$

Step 5.3.5

Write $1$ as a fraction with a common denominator.

$\sqrt{\frac{25}{25} - \frac{1}{25}}$

Step 5.3.6

Combine the numerators over the common denominator.

$\sqrt{\frac{25 - 1}{25}}$

Step 5.3.7

Subtract $1$ from $25$ .

$\sqrt{\frac{24}{25}}$

Step 5.3.8

Rewrite $\sqrt{\frac{24}{25}}$ as $\frac{\sqrt{24}}{\sqrt{25}}$ .

$\frac{\sqrt{24}}{\sqrt{25}}$

Step 5.3.9

Simplify the numerator.

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Step 5.3.9.1

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 5.3.9.1.1

Factor $4$ out of $24$ .

$\frac{\sqrt{4 (6)}}{\sqrt{25}}$

Step 5.3.9.1.2

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

$\frac{\sqrt{2^{2} \cdot 6}}{\sqrt{25}}$

Step 5.3.9.2

Pull terms out from under the radical.

$\frac{2 \sqrt{6}}{\sqrt{25}}$

Step 5.3.10

Simplify the denominator.

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Step 5.3.10.1

Rewrite $25$ as $5^{2}$ .

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

Step 5.3.10.2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 6

Find the vertices.

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Step 6.1

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

$(h, k + a)$

Step 6.2

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

$(0, 0 + 1)$

Step 6.3

Simplify.

$(0, 1)$

Step 6.4

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

$(h, k - a)$

Step 6.5

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

$(0, 0 - (1))$

Step 6.6

Simplify.

$(0, - 1)$

Step 6.7

Ellipses have two vertices.

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

${Vertex}_{2}$ : $(0, - 1)$

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

${Vertex}_{2}$ : $(0, - 1)$

Step 7

Find the foci.

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Step 7.1

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

$(h, k + c)$

Step 7.2

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

$(0, 0 + \frac{2 \sqrt{6}}{5})$

Step 7.3

Simplify.

$(0, \frac{2 \sqrt{6}}{5})$

Step 7.4

The first focus of an ellipse can be found by subtracting $c$ from $k$ .

$(h, k - c)$

Step 7.5

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

$(0, 0 - (\frac{2 \sqrt{6}}{5}))$

Step 7.6

Simplify.

$(0, - \frac{2 \sqrt{6}}{5})$

Step 7.7

Ellipses have two foci.

${Focus}_{1}$ : $(0, \frac{2 \sqrt{6}}{5})$

${Focus}_{2}$ : $(0, - \frac{2 \sqrt{6}}{5})$

${Focus}_{1}$ : $(0, \frac{2 \sqrt{6}}{5})$

${Focus}_{2}$ : $(0, - \frac{2 \sqrt{6}}{5})$

Step 8

Find the eccentricity.

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Step 8.1

Find the eccentricity by using the following formula.

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

Step 8.2

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

$\frac{\sqrt{{(1)}^{2} - {(\frac{1}{5})}^{2}}}{1}$

Step 8.3

Simplify.

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Step 8.3.1

Divide $\sqrt{{(1)}^{2} - {(\frac{1}{5})}^{2}}$ by $1$ .

$\sqrt{{(1)}^{2} - {(\frac{1}{5})}^{2}}$

Step 8.3.2

One to any power is one.

$\sqrt{1 - {(\frac{1}{5})}^{2}}$

Step 8.3.3

Apply the product rule to $\frac{1}{5}$ .

$\sqrt{1 - \frac{1^{2}}{5^{2}}}$

Step 8.3.4

One to any power is one.

$\sqrt{1 - \frac{1}{5^{2}}}$

Step 8.3.5

Raise $5$ to the power of $2$ .

$\sqrt{1 - \frac{1}{25}}$

Step 8.3.6

Write $1$ as a fraction with a common denominator.

$\sqrt{\frac{25}{25} - \frac{1}{25}}$

Step 8.3.7

Combine the numerators over the common denominator.

$\sqrt{\frac{25 - 1}{25}}$

Step 8.3.8

Subtract $1$ from $25$ .

$\sqrt{\frac{24}{25}}$

Step 8.3.9

Rewrite $\sqrt{\frac{24}{25}}$ as $\frac{\sqrt{24}}{\sqrt{25}}$ .

$\frac{\sqrt{24}}{\sqrt{25}}$

Step 8.3.10

Simplify the numerator.

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Step 8.3.10.1

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 8.3.10.1.1

Factor $4$ out of $24$ .

$\frac{\sqrt{4 (6)}}{\sqrt{25}}$

Step 8.3.10.1.2

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

$\frac{\sqrt{2^{2} \cdot 6}}{\sqrt{25}}$

Step 8.3.10.2

Pull terms out from under the radical.

$\frac{2 \sqrt{6}}{\sqrt{25}}$

Step 8.3.11

Simplify the denominator.

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Step 8.3.11.1

Rewrite $25$ as $5^{2}$ .

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

Step 8.3.11.2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 9

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

Center: $(0, 0)$

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

${Vertex}_{2}$ : $(0, - 1)$

${Focus}_{1}$ : $(0, \frac{2 \sqrt{6}}{5})$

${Focus}_{2}$ : $(0, - \frac{2 \sqrt{6}}{5})$

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

Step 10