Find the Center 3x^2-4y^2-6x-16y+7=0

$3 x^{2} - 4 y^{2} - 6 x - 16 y + 7 = 0$

Step 1

Find the standard form of the hyperbola.

Subtract $7$ from both sides of the equation.

$3 x^{2} - 4 y^{2} - 6 x - 16 y = - 7$

Complete the square for $3 x^{2} - 6 x$ .

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

$a = 3$

$b = - 6$

$c = 0$

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

$d = \frac{- 6}{2 \cdot 3}$

Simplify the right side.

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

$d = \frac{2 \cdot - 3}{2 \cdot 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 3$ .

$d = \frac{2 \cdot - 3}{2 (3)}$

Cancel the common factor.

$d = \frac{2 \cdot - 3}{2 \cdot 3}$

Rewrite the expression.

$d = \frac{- 3}{3}$

Cancel the common factor of $- 3$ and $3$ .

Factor $3$ out of $- 3$ .

$d = \frac{3 \cdot - 1}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$d = \frac{3 \cdot - 1}{3 (1)}$

Cancel the common factor.

$d = \frac{3 \cdot - 1}{3 \cdot 1}$

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$d = - 1$

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

$e = 0 - \frac{{(- 6)}^{2}}{4 \cdot 3}$

Simplify the right side.

Simplify each term.

Raise $- 6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot 3}$

Multiply $4$ by $3$ .

$e = 0 - \frac{36}{12}$

Divide $36$ by $12$ .

$e = 0 - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$e = 0 - 3$

Subtract $3$ from $0$ .

$e = - 3$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(x - 1)}^{2} - 3$ .

$3 {(x - 1)}^{2} - 3$

Substitute $3 {(x - 1)}^{2} - 3$ for $3 x^{2} - 6 x$ in the equation $3 x^{2} - 4 y^{2} - 6 x - 16 y = - 7$ .

$3 {(x - 1)}^{2} - 3 - 4 y^{2} - 16 y = - 7$

Move $- 3$ to the right side of the equation by adding $3$ to both sides.

$3 {(x - 1)}^{2} - 4 y^{2} - 16 y = - 7 + 3$

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 $- 8$ .

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

Cancel the common factors.

Factor $- 4$ out of $- 4$ .

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

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 $- 16$ .

$e = 0 - - 16$

Multiply $- 1$ by $- 16$ .

$e = 0 + 16$

Add $0$ and $16$ .

$e = 16$

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 $3 x^{2} - 4 y^{2} - 6 x - 16 y = - 7$ .

$3 {(x - 1)}^{2} - 4 {(y + 2)}^{2} + 16 = - 7 + 3$

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

$3 {(x - 1)}^{2} - 4 {(y + 2)}^{2} = - 7 + 3 - 16$

Simplify $- 7 + 3 - 16$ .

Add $- 7$ and $3$ .

$3 {(x - 1)}^{2} - 4 {(y + 2)}^{2} = - 4 - 16$

Subtract $16$ from $- 4$ .

$3 {(x - 1)}^{2} - 4 {(y + 2)}^{2} = - 20$

Flip the sign on each term of the equation so the term on the right side is positive.

$- 3 {(x - 1)}^{2} + 4 {(y + 2)}^{2} = 20$

Divide each term by $20$ to make the right side equal to one.

$- \frac{3 {(x - 1)}^{2}}{20} + \frac{4 {(y + 2)}^{2}}{20} = \frac{20}{20}$

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 $1$ .

$\frac{{(y + 2)}^{2}}{5} - \frac{{(x - 1)}^{2}}{\frac{20}{3}} = 1$

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, $a$ .

$a = \sqrt{5}$

$b = \frac{2 \sqrt{15}}{3}$

$k = - 2$

$h = 1$

Step 4

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

$(1, - 2)$

Step 5

Find the Center 3x^2-4y^2-6x-16y+7=0

Related Questions

Support LCMGCF.com by disabling your adblocker.