Graph (x-1)^2-y^2=4

${(x - 1)}^{2} - y^{2} = 4$

Find the standard form of the hyperbola.

Divide each term by $4$ to make the right side equal to one.

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

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 $1$ .

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

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

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

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, $a$ .

$a = 2$

$b = 2$

$k = 0$

$h = 1$

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(1, 0)$

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

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

$\sqrt{{(2)}^{2} + {(2)}^{2}}$

Simplify.

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

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

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

$\sqrt{4 + 4}$

Add $4$ and $4$ .

$\sqrt{8}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$\sqrt{4 (2)}$

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

$\sqrt{2^{2} \cdot 2}$

Pull terms out from under the radical.

$2 \sqrt{2}$

Find the vertices.

The first vertex of a hyperbola can be found by adding $a$ to $h$ .

$(h + a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(3, 0)$

The second vertex of a hyperbola can be found by subtracting $a$ from $h$ .

$(h - a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(- 1, 0)$

The vertices of a hyperbola follow the form of $(h \pm a, k)$ . Hyperbolas have two vertices.

$(3, 0), (- 1, 0)$

Find the foci.

The first focus of a hyperbola can be found by adding $c$ to $h$ .

$(h + c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(1 + 2 \sqrt{2}, 0)$

The second focus of a hyperbola can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(1 - 2 \sqrt{2}, 0)$

The foci of a hyperbola follow the form of $(h \pm \sqrt{a^{2} + b^{2}}, k)$ . Hyperbolas have two foci.

$(1 + 2 \sqrt{2}, 0), (1 - 2 \sqrt{2}, 0)$

Find the eccentricity.

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.

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

Simplify.

Simplify the numerator.

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

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

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

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

Add $4$ and $4$ .

$\frac{\sqrt{8}}{2}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

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

Pull terms out from under the radical.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $\sqrt{2}$ by $1$ .

$\sqrt{2}$

The asymptotes follow the form $y = \pm \frac{b (x - h)}{a} + k$ because this hyperbola opens left and right.

$y = \pm 1 \cdot (x - (1)) + 0$

Simplify to find the first asymptote.

Remove parentheses.

$y = 1 \cdot (x - (1)) + 0$

Simplify $1 \cdot (x - (1)) + 0$ .

Add $1 \cdot (x - (1))$ and $0$ .

$y = 1 \cdot (x - (1))$

Multiply $x - (1)$ by $1$ .

$y = x - (1)$

Multiply $- 1$ by $1$ .

$y = x - 1$

Simplify to find the second asymptote.

Remove parentheses.

$y = - 1 \cdot (x - (1)) + 0$

Simplify $- 1 \cdot (x - (1)) + 0$ .

Simplify the expression.

Add $- 1 \cdot (x - (1))$ and $0$ .

$y = - 1 \cdot (x - (1))$

Multiply $- 1$ by $1$ .

$y = - 1 \cdot (x - 1)$

Apply the distributive property.

$y = - 1 x - 1 \cdot - 1$

Simplify the expression.

Rewrite $- 1 x$ as $- x$ .

$y = - x - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$y = - x + 1$

This hyperbola has two asymptotes.

$y = x - 1, y = - x + 1$

These values represent the important values for graphing and analyzing a hyperbola.

Center: $(1, 0)$

Vertices: $(3, 0), (- 1, 0)$

Foci: $(1 + 2 \sqrt{2}, 0), (1 - 2 \sqrt{2}, 0)$

Eccentricity: $\sqrt{2}$

Focal Parameter: $\sqrt{2}$

Asymptotes: $y = x - 1$ , $y = - x + 1$

Graph (x-1)^2-y^2=4

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