Find the Focus -x^2+4y^2-16y+12=0

$- x^{2} + 4 y^{2} - 16 y + 12 = 0$

Step 1

Find the standard form of the hyperbola.

Subtract $12$ from both sides of the equation.

$- x^{2} + 4 y^{2} - 16 y = - 12$

Complete the square for $4 y^{2} - 16 y$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 4$

$b = - 16$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 16}{2 \cdot 4}$

Simplify the right side.

Cancel the common factor of $- 16$ and $2$ .

Factor $2$ out of $- 16$ .

$d = \frac{2 \cdot - 8}{2 \cdot 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 4$ .

$d = \frac{2 \cdot - 8}{2 (4)}$

Cancel the common factor.

$d = \frac{2 \cdot - 8}{2 \cdot 4}$

Rewrite the expression.

$d = \frac{- 8}{4}$

Cancel the common factor of $- 8$ and $4$ .

Factor $4$ out of $- 8$ .

$d = \frac{4 \cdot - 2}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$d = \frac{4 \cdot - 2}{4 (1)}$

Cancel the common factor.

$d = \frac{4 \cdot - 2}{4 \cdot 1}$

Rewrite the expression.

$d = \frac{- 2}{1}$

Divide $- 2$ by $1$ .

$d = - 2$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 16)}^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

Raise $- 16$ to the power of $2$ .

$e = 0 - \frac{256}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{256}{16}$

Divide $256$ by $16$ .

$e = 0 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Subtract $16$ from $0$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(y - 2)}^{2} - 16$ .

$4 {(y - 2)}^{2} - 16$

Substitute $4 {(y - 2)}^{2} - 16$ for $4 y^{2} - 16 y$ in the equation $- x^{2} + 4 y^{2} - 16 y = - 12$ .

$- x^{2} + 4 {(y - 2)}^{2} - 16 = - 12$

Move $- 16$ to the right side of the equation by adding $16$ to both sides.

$- x^{2} + 4 {(y - 2)}^{2} = - 12 + 16$

Add $- 12$ and $16$ .

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

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

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

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 $1$ .

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

Step 2

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

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

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, $a$ .

$a = 1$

$b = 2$

$k = 2$

$h = 0$

Step 4

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.

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

Simplify.

One to any power is one.

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

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

$\sqrt{1 + 4}$

Add $1$ and $4$ .

$\sqrt{5}$

Step 5

Find the foci.

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

$(h, k + c)$

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

$(0, 2 + \sqrt{5})$

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

$(h, k - c)$

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

$(0, 2 - \sqrt{5})$

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

$(0, 2 + \sqrt{5}), (0, 2 - \sqrt{5})$

Step 6

Find the Focus -x^2+4y^2-16y+12=0

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