Find the Eccentricity ((x+1)^2)/25-((y-4)^2)/16=1

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

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 $1$ .

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

This is the standard form of a hyperbola. Use this form to determine the eccentricity.

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

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, $a$ .

$a = 5$

$b = 4$

$k = 4$

$h = - 1$

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.

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

Simplify the numerator.

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

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

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

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

Add $25$ and $16$ .

$\frac{\sqrt{41}}{5}$

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{41}}{5}$

Decimal Form:

$1.28062484 \dots$

Find the Eccentricity ((x+1)^2)/25-((y-4)^2)/16=1

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