Find the Eccentricity 16x^2-y^2=1

$16 x^{2} - y^{2} = 1$

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 $1$ .

$\frac{x^{2}}{\frac{1}{16}} - \frac{y^{2}}{1} = 1$

Step 2

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

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

Step 3

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, $a$ .

$a = \frac{1}{4}$

$b = 1$

$k = 0$

$h = 0$

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.

$\frac{\sqrt{{(\frac{1}{4})}^{2} + {(1)}^{2}}}{\frac{1}{4}}$

Step 6

Simplify.

Multiply the numerator by the reciprocal of the denominator.

$\sqrt{{(\frac{1}{4})}^{2} + 1^{2}} \cdot 4$

Apply the product rule to $\frac{1}{4}$ .

$\sqrt{\frac{1^{2}}{4^{2}} + 1^{2}} \cdot 4$

One to any power is one.

$\sqrt{\frac{1}{4^{2}} + 1^{2}} \cdot 4$

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

$\sqrt{\frac{1}{16} + 1^{2}} \cdot 4$

One to any power is one.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $1$ and $16$ .

$\sqrt{\frac{17}{16}} \cdot 4$

Rewrite $\sqrt{\frac{17}{16}}$ as $\frac{\sqrt{17}}{\sqrt{16}}$ .

$\frac{\sqrt{17}}{\sqrt{16}} \cdot 4$

Simplify the denominator.

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

$\frac{\sqrt{17}}{\sqrt{4^{2}}} \cdot 4$

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

$\frac{\sqrt{17}}{4} \cdot 4$

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{\sqrt{17}}{4} \cdot 4$

Rewrite the expression.

$\sqrt{17}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\sqrt{17}$

Decimal Form:

$4.12310562 \dots$

Step 8

Find the Eccentricity 16x^2-y^2=1

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