Precalculus Examples

$\frac{{(x + 4)}^{2}}{25} + \frac{{(y - 1)}^{2}}{16} = 1$

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 $1$ .

$\frac{{(x + 4)}^{2}}{25} + \frac{{(y - 1)}^{2}}{16} = 1$

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.

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

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.

$a = 5$

$b = 4$

$k = 1$

$h = - 4$

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.

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

Step 6

Simplify the numerator.

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Step 6.1

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

$\frac{\sqrt{25 - 4^{2}}}{5}$

Step 6.2

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

$\frac{\sqrt{25 - 1 \cdot 16}}{5}$

Step 6.3

Multiply $- 1$ by $16$ .

$\frac{\sqrt{25 - 16}}{5}$

Step 6.4

Subtract $16$ from $25$ .

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

Step 6.5

Rewrite $9$ as $3^{2}$ .

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

Step 6.6

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

$\frac{3}{5}$

Step 7