Algebra Examples

$\frac{{(x - 5)}^{2}}{4} - \frac{{(y - 2)}^{2}}{16} = 1$

Step 1

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 $1$ .

$\frac{{(x - 5)}^{2}}{4} - \frac{{(y - 2)}^{2}}{16} = 1$

Step 2

This is the standard form of a hyperbola. Use this form to determine the eccentricity.

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

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 2$

$b = 4$

$k = 2$

$h = 5$

Step 4

Find the eccentricity by using the following formula.

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

Step 5

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

$\frac{\sqrt{{(2)}^{2} + {(4)}^{2}}}{2}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

$\frac{\sqrt{4 + 4^{2}}}{2}$

Step 6.1.2

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

$\frac{\sqrt{4 + 16}}{2}$

Step 6.1.3

Add $4$ and $16$ .

$\frac{\sqrt{20}}{2}$

Step 6.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 6.1.4.1

Factor $4$ out of $20$ .

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

Step 6.1.4.2

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

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

Step 6.1.5

Pull terms out from under the radical.

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

Step 6.2

Cancel the common factor of $2$ .

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Step 6.2.1

Cancel the common factor.

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

Step 6.2.2

Divide $\sqrt{5}$ by $1$ .

$\sqrt{5}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\sqrt{5}$

Decimal Form:

$2.23606797 \dots$

Step 8