Find the Eccentricity 25x^2+9y^2=225

$25 x^{2} + 9 y^{2} = 225$

Divide each term by $225$ to make the right side equal to one.

$\frac{25 x^{2}}{225} + \frac{9 y^{2}}{225} = \frac{225}{225}$

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 $1$ .

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

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.

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

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.

$a = 5$

$b = 3$

$k = 0$

$h = 0$

Find the eccentricity by using the following formula.

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

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

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

Simplify the numerator.

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

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

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

$\frac{\sqrt{25 - 1 \cdot 9}}{5}$

Multiply $- 1$ by $9$ .

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

Subtract $9$ from $25$ .

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

Rewrite $16$ as $4^{2}$ .

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

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

$\frac{4}{5}$

Find the Eccentricity 25x^2+9y^2=225

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