Trigonometry Examples

$4 x^{2} + y^{2} = 144$

Step 1

Find the standard form of the ellipse.

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Step 1.1

Divide each term by $144$ to make the right side equal to one.

$\frac{4 x^{2}}{144} + \frac{y^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{144} = 1$

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.

$a = 12$

$b = 6$

$k = 0$

$h = 0$

Step 4

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

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Step 4.1

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

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

Step 4.2

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

$\sqrt{{(12)}^{2} - {(6)}^{2}}$

Step 4.3

Simplify.

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Step 4.3.1

Raise $12$ to the power of $2$ .

$\sqrt{144 - {(6)}^{2}}$

Step 4.3.2

Raise $6$ to the power of $2$ .

$\sqrt{144 - 1 \cdot 36}$

Step 4.3.3

Multiply $- 1$ by $36$ .

$\sqrt{144 - 36}$

Step 4.3.4

Subtract $36$ from $144$ .

$\sqrt{108}$

Step 4.3.5

Rewrite $108$ as $6^{2} \cdot 3$ .

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Step 4.3.5.1

Factor $36$ out of $108$ .

$\sqrt{36 (3)}$

Step 4.3.5.2

Rewrite $36$ as $6^{2}$ .

$\sqrt{6^{2} \cdot 3}$

Step 4.3.6

Pull terms out from under the radical.

$6 \sqrt{3}$

Step 5

Find the foci.

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Step 5.1

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

$(h, k + c)$

Step 5.2

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

$(0, 0 + 6 \sqrt{3})$

Step 5.3

Simplify.

$(0, 6 \sqrt{3})$

Step 5.4

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

$(h, k - c)$

Step 5.5

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

$(0, 0 - (6 \sqrt{3}))$

Step 5.6

Simplify.

$(0, - 6 \sqrt{3})$

Step 5.7

Ellipses have two foci.

${Focus}_{1}$ : $(0, 6 \sqrt{3})$

${Focus}_{2}$ : $(0, - 6 \sqrt{3})$

${Focus}_{1}$ : $(0, 6 \sqrt{3})$

${Focus}_{2}$ : $(0, - 6 \sqrt{3})$

Step 6