Find the Eccentricity 9x^2+4y^2=36

$9 x^{2} + 4 y^{2} = 36$

Divide each term by $36$ to make the right side equal to one.

$\frac{9 x^{2}}{36} + \frac{4 y^{2}}{36} = \frac{36}{36}$

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 $1$ .

$\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

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.

$a = 3$

$b = 2$

$k = 0$

$h = 0$

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.

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

Simplify the numerator.

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

$\frac{\sqrt{9 - 2^{2}}}{3}$

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

$\frac{\sqrt{9 - 1 \cdot 4}}{3}$

Multiply $- 1$ by $4$ .

$\frac{\sqrt{9 - 4}}{3}$

Subtract $4$ from $9$ .

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.74535599 \dots$

Find the Eccentricity 9x^2+4y^2=36

Related Questions

Support LCMGCF.com by disabling your adblocker.