Precalculus Examples

x^{2} + 4 y^{2} = 1

$x^{2} + 4 y^{2} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to

1

$1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be

1

$1$ .

x^{2} + \frac{y^{2}}{\frac{1}{4}} = 1

$x^{2} + \frac{y^{2}}{\frac{1}{4}} = 1$

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

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

Step 3

Match the values in this ellipse to those of the standard form. The variable

a

$a$ represents the radius of the major axis of the ellipse,

b

$b$ represents the radius of the minor axis of the ellipse,

h

$h$ represents the x-offset from the origin, and

k

$k$ represents the y-offset from the origin.

a = 1

$a = 1$

b = \frac{1}{2}

$b = \frac{1}{2}$

k = 0

$k = 0$

h = 0

$h = 0$

Step 4

The center of an ellipse follows the form of

(h, k)

$(h, k)$ . Substitute in the values of

h

$h$ and

k

$k$ .

(0, 0)

$(0, 0)$

Step 5

Find

c

$c$ , 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}}

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

Step 5.2

Substitute the values of

a

$a$ and

b

$b$ in the formula.

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

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

Step 5.3

Simplify.

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Step 5.3.1

One to any power is one.

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

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

Step 5.3.2

Apply the product rule to

\frac{1}{2}

$\frac{1}{2}$ .

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

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

Step 5.3.3

One to any power is one.

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

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

Step 5.3.4

Raise

2

$2$ to the power of

2

$2$ .

\sqrt{1 - \frac{1}{4}}

$\sqrt{1 - \frac{1}{4}}$

Step 5.3.5

Write

1

$1$ as a fraction with a common denominator.

\sqrt{\frac{4}{4} - \frac{1}{4}}

$\sqrt{\frac{4}{4} - \frac{1}{4}}$

Step 5.3.6

Combine the numerators over the common denominator.

\sqrt{\frac{4 - 1}{4}}

$\sqrt{\frac{4 - 1}{4}}$

Step 5.3.7

Subtract

1

$1$ from

4

$4$ .

\sqrt{\frac{3}{4}}

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

Step 5.3.8

Rewrite

\sqrt{\frac{3}{4}}

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

\frac{\sqrt{3}}{\sqrt{4}}

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

\frac{\sqrt{3}}{\sqrt{4}}

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

Step 5.3.9

Simplify the denominator.

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Step 5.3.9.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

\frac{\sqrt{3}}{\sqrt{2^{2}}}

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

Step 5.3.9.2

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

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

Step 6

Find the vertices.

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Step 6.1

The first vertex of an ellipse can be found by adding

a

$a$ to

h

$h$ .

(h + a, k)

$(h + a, k)$

Step 6.2

Substitute the known values of

h

$h$ ,

a

$a$ , and

k

$k$ into the formula.

(0 + 1, 0)

$(0 + 1, 0)$

Step 6.3

Simplify.

(1, 0)

$(1, 0)$

Step 6.4

The second vertex of an ellipse can be found by subtracting

a

$a$ from

h

$h$ .

(h - a, k)

$(h - a, k)$

Step 6.5

Substitute the known values of

h

$h$ ,

a

$a$ , and

k

$k$ into the formula.

(0 - (1), 0)

$(0 - (1), 0)$

Step 6.6

Simplify.

(- 1, 0)

$(- 1, 0)$

Step 6.7

Ellipses have two vertices.

{Vertex}_{1}

${Vertex}_{1}$ :

(1, 0)

$(1, 0)$

{Vertex}_{2}

${Vertex}_{2}$ :

(- 1, 0)

$(- 1, 0)$

{Vertex}_{1}

${Vertex}_{1}$ :

(1, 0)

$(1, 0)$

{Vertex}_{2}

${Vertex}_{2}$ :

(- 1, 0)

$(- 1, 0)$

Step 7

Find the foci.

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Step 7.1

The first focus of an ellipse can be found by adding

c

$c$ to

h

$h$ .

(h + c, k)

$(h + c, k)$

Step 7.2

Substitute the known values of

h

$h$ ,

c

$c$ , and

k

$k$ into the formula.

(0 + \frac{\sqrt{3}}{2}, 0)

$(0 + \frac{\sqrt{3}}{2}, 0)$

Step 7.3

Simplify.

(\frac{\sqrt{3}}{2}, 0)

$(\frac{\sqrt{3}}{2}, 0)$

Step 7.4

The second focus of an ellipse can be found by subtracting

c

$c$ from

h

$h$ .

(h - c, k)

$(h - c, k)$

Step 7.5

Substitute the known values of

h

$h$ ,

c

$c$ , and

k

$k$ into the formula.

(0 - (\frac{\sqrt{3}}{2}), 0)

$(0 - (\frac{\sqrt{3}}{2}), 0)$

Step 7.6

Simplify.

(- \frac{\sqrt{3}}{2}, 0)

$(- \frac{\sqrt{3}}{2}, 0)$

Step 7.7

Ellipses have two foci.

{Focus}_{1}

${Focus}_{1}$ :

(\frac{\sqrt{3}}{2}, 0)

$(\frac{\sqrt{3}}{2}, 0)$

{Focus}_{2}

${Focus}_{2}$ :

(- \frac{\sqrt{3}}{2}, 0)

$(- \frac{\sqrt{3}}{2}, 0)$

{Focus}_{1}

${Focus}_{1}$ :

(\frac{\sqrt{3}}{2}, 0)

$(\frac{\sqrt{3}}{2}, 0)$

{Focus}_{2}

${Focus}_{2}$ :

(- \frac{\sqrt{3}}{2}, 0)

$(- \frac{\sqrt{3}}{2}, 0)$

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}

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

Step 8.2

Substitute the values of

a

$a$ and

b

$b$ into the formula.

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

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

Step 8.3

Simplify.

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Step 8.3.1

Divide

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

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

1

$1$ .

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

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

Step 8.3.2

One to any power is one.

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

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

Step 8.3.3

Apply the product rule to

\frac{1}{2}

$\frac{1}{2}$ .

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

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

Step 8.3.4

One to any power is one.

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

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

Step 8.3.5

Raise

2

$2$ to the power of

2

$2$ .

\sqrt{1 - \frac{1}{4}}

$\sqrt{1 - \frac{1}{4}}$

Step 8.3.6

Write

1

$1$ as a fraction with a common denominator.

\sqrt{\frac{4}{4} - \frac{1}{4}}

$\sqrt{\frac{4}{4} - \frac{1}{4}}$

Step 8.3.7

Combine the numerators over the common denominator.

\sqrt{\frac{4 - 1}{4}}

$\sqrt{\frac{4 - 1}{4}}$

Step 8.3.8

Subtract

1

$1$ from

4

$4$ .

\sqrt{\frac{3}{4}}

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

Step 8.3.9

Rewrite

\sqrt{\frac{3}{4}}

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

\frac{\sqrt{3}}{\sqrt{4}}

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

\frac{\sqrt{3}}{\sqrt{4}}

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

Step 8.3.10

Simplify the denominator.

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Step 8.3.10.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

\frac{\sqrt{3}}{\sqrt{2^{2}}}

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

Step 8.3.10.2

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

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

Step 9

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

Center:

(0, 0)

$(0, 0)$

{Vertex}_{1}

${Vertex}_{1}$ :

(1, 0)

$(1, 0)$

{Vertex}_{2}

${Vertex}_{2}$ :

(- 1, 0)

$(- 1, 0)$

{Focus}_{1}

${Focus}_{1}$ :

(\frac{\sqrt{3}}{2}, 0)

$(\frac{\sqrt{3}}{2}, 0)$

{Focus}_{2}

${Focus}_{2}$ :

(- \frac{\sqrt{3}}{2}, 0)

$(- \frac{\sqrt{3}}{2}, 0)$

Eccentricity:

\frac{\sqrt{3}}{2}

$\frac{\sqrt{3}}{2}$

Step 10

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