Find the Foci 9y^2-72y-16x^2-64x-64=0

$9 y^{2} - 72 y - 16 x^{2} - 64 x - 64 = 0$

Step 1

Find the standard form of the hyperbola.

Add $64$ to both sides of the equation.

$9 y^{2} - 72 y - 16 x^{2} - 64 x = 64$

Complete the square for $9 y^{2} - 72 y$ .

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

$a = 9$

$b = - 72$

$c = 0$

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 $d = \frac{b}{2 a}$ .

$d = \frac{- 72}{2 \cdot 9}$

Simplify the right side.

Cancel the common factor of $- 72$ and $2$ .

Factor $2$ out of $- 72$ .

$d = \frac{2 \cdot - 36}{2 \cdot 9}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 9$ .

$d = \frac{2 \cdot - 36}{2 (9)}$

Cancel the common factor.

$d = \frac{2 \cdot - 36}{2 \cdot 9}$

Rewrite the expression.

$d = \frac{- 36}{9}$

Cancel the common factor of $- 36$ and $9$ .

Factor $9$ out of $- 36$ .

$d = \frac{9 \cdot - 4}{9}$

Cancel the common factors.

Factor $9$ out of $9$ .

$d = \frac{9 \cdot - 4}{9 (1)}$

Cancel the common factor.

$d = \frac{9 \cdot - 4}{9 \cdot 1}$

Rewrite the expression.

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

Divide $- 4$ by $1$ .

$d = - 4$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 72)}^{2}}{4 \cdot 9}$

Simplify the right side.

Simplify each term.

Raise $- 72$ to the power of $2$ .

$e = 0 - \frac{5184}{4 \cdot 9}$

Multiply $4$ by $9$ .

$e = 0 - \frac{5184}{36}$

Divide $5184$ by $36$ .

$e = 0 - 1 \cdot 144$

Multiply $- 1$ by $144$ .

$e = 0 - 144$

Subtract $144$ from $0$ .

$e = - 144$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $9 {(y - 4)}^{2} - 144$ .

$9 {(y - 4)}^{2} - 144$

Substitute $9 {(y - 4)}^{2} - 144$ for $9 y^{2} - 72 y$ in the equation $9 y^{2} - 72 y - 16 x^{2} - 64 x = 64$ .

$9 {(y - 4)}^{2} - 144 - 16 x^{2} - 64 x = 64$

Move $- 144$ to the right side of the equation by adding $144$ to both sides.

$9 {(y - 4)}^{2} - 16 x^{2} - 64 x = 64 + 144$

Complete the square for $- 16 x^{2} - 64 x$ .

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

$a = - 16$

$b = - 64$

$c = 0$

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 $d = \frac{b}{2 a}$ .

$d = \frac{- 64}{2 \cdot - 16}$

Simplify the right side.

Cancel the common factor of $- 64$ and $2$ .

Factor $2$ out of $- 64$ .

$d = \frac{2 \cdot - 32}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$d = \frac{2 \cdot - 32}{2 (- 16)}$

Cancel the common factor.

$d = \frac{2 \cdot - 32}{2 \cdot - 16}$

Rewrite the expression.

$d = \frac{- 32}{- 16}$

Cancel the common factor of $- 32$ and $- 16$ .

Factor $- 16$ out of $- 32$ .

$d = \frac{- 16 (2)}{- 16}$

Cancel the common factors.

Factor $- 16$ out of $- 16$ .

$d = \frac{- 16 \cdot 2}{- 16 \cdot 1}$

Cancel the common factor.

$d = \frac{- 16 \cdot 2}{- 16 \cdot 1}$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 64)}^{2}}{4 \cdot - 16}$

Simplify the right side.

Simplify each term.

Raise $- 64$ to the power of $2$ .

$e = 0 - \frac{4096}{4 \cdot - 16}$

Multiply $4$ by $- 16$ .

$e = 0 - \frac{4096}{- 64}$

Divide $4096$ by $- 64$ .

$e = 0 - - 64$

Multiply $- 1$ by $- 64$ .

$e = 0 + 64$

Add $0$ and $64$ .

$e = 64$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 16 {(x + 2)}^{2} + 64$ .

$- 16 {(x + 2)}^{2} + 64$

Substitute $- 16 {(x + 2)}^{2} + 64$ for $- 16 x^{2} - 64 x$ in the equation $9 y^{2} - 72 y - 16 x^{2} - 64 x = 64$ .

$9 {(y - 4)}^{2} - 16 {(x + 2)}^{2} + 64 = 64 + 144$

Move $64$ to the right side of the equation by adding $64$ to both sides.

$9 {(y - 4)}^{2} - 16 {(x + 2)}^{2} = 64 + 144 - 64$

Simplify $64 + 144 - 64$ .

Add $64$ and $144$ .

$9 {(y - 4)}^{2} - 16 {(x + 2)}^{2} = 208 - 64$

Subtract $64$ from $208$ .

$9 {(y - 4)}^{2} - 16 {(x + 2)}^{2} = 144$

Divide each term by $144$ to make the right side equal to one.

$\frac{9 {(y - 4)}^{2}}{144} - \frac{16 {(x + 2)}^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{{(y - 4)}^{2}}{16} - \frac{{(x + 2)}^{2}}{9} = 1$

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, $a$ .

$a = 4$

$b = 3$

$k = 4$

$h = - 2$

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.

$\sqrt{{(4)}^{2} + {(3)}^{2}}$

Simplify.

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

$\sqrt{16 + {(3)}^{2}}$

Raise $3$ to the power of $2$ .

$\sqrt{16 + 9}$

Add $16$ and $9$ .

$\sqrt{25}$

Rewrite $25$ as $5^{2}$ .

$\sqrt{5^{2}}$

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

$5$

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.

$(- 2, 9)$

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.

$(- 2, - 1)$

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

$(- 2, 9), (- 2, - 1)$

Step 6

Find the Foci 9y^2-72y-16x^2-64x-64=0

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