Precalculus Examples

$\frac{{(x + 5)}^{2}}{25} - \frac{{(y - 5)}^{2}}{9} = 1$

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 $1$ .

$\frac{{(x + 5)}^{2}}{25} - \frac{{(y - 5)}^{2}}{9} = 1$

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, $a$ .

$a = 5$

$b = 3$

$k = 5$

$h = - 5$

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.

$\frac{\sqrt{{(5)}^{2} + {(3)}^{2}}}{5}$

Step 6

Simplify the numerator.

Tap for more steps...

Step 6.1

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

$\frac{\sqrt{25 + 3^{2}}}{5}$

Step 6.2

Raise $3$ to the power of $2$ .

$\frac{\sqrt{25 + 9}}{5}$

Step 6.3

Add $25$ and $9$ .

$\frac{\sqrt{34}}{5}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{34}}{5}$

Decimal Form:

$1.16619037 \dots$

Step 8