Find the Eccentricity ((x-2)^2)/16+((y-1)^2)/4=1

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

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 $1$ .

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

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.

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

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.

$a = 4$

$b = 2$

$k = 1$

$h = 2$

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.

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

Simplify.

Simplify the numerator.

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

$\frac{\sqrt{16 - 2^{2}}}{4}$

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

$\frac{\sqrt{16 - 1 \cdot 4}}{4}$

Multiply $- 1$ by $4$ .

$\frac{\sqrt{16 - 4}}{4}$

Subtract $4$ from $16$ .

$\frac{\sqrt{12}}{4}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$\frac{\sqrt{4 (3)}}{4}$

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

$\frac{\sqrt{2^{2} \cdot 3}}{4}$

Pull terms out from under the radical.

$\frac{2 \sqrt{3}}{4}$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2 \sqrt{3}$ .

$\frac{2 (\sqrt{3})}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$\frac{2 \sqrt{3}}{2 \cdot 2}$

Cancel the common factor.

$\frac{2 \sqrt{3}}{2 \cdot 2}$

Rewrite the expression.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.86602540 \dots$

Find the Eccentricity ((x-2)^2)/16+((y-1)^2)/4=1

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