Find the Asymptotes 81x^2-y^2-810x-10y+2009=0

$81 x^{2} - y^{2} - 810 x - 10 y + 2009 = 0$

Step 1

Find the standard form of the hyperbola.

Subtract $2009$ from both sides of the equation.

$81 x^{2} - y^{2} - 810 x - 10 y = - 2009$

Complete the square for $81 x^{2} - 810 x$ .

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

$a = 81$

$b = - 810$

$c = 0$

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

$d = \frac{- 810}{2 \cdot 81}$

Simplify the right side.

Cancel the common factor of $- 810$ and $2$ .

Factor $2$ out of $- 810$ .

$d = \frac{2 \cdot - 405}{2 \cdot 81}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 81$ .

$d = \frac{2 \cdot - 405}{2 (81)}$

Cancel the common factor.

$d = \frac{2 \cdot - 405}{2 \cdot 81}$

Rewrite the expression.

$d = \frac{- 405}{81}$

Cancel the common factor of $- 405$ and $81$ .

Factor $81$ out of $- 405$ .

$d = \frac{81 \cdot - 5}{81}$

Cancel the common factors.

Factor $81$ out of $81$ .

$d = \frac{81 \cdot - 5}{81 (1)}$

Cancel the common factor.

$d = \frac{81 \cdot - 5}{81 \cdot 1}$

Rewrite the expression.

$d = \frac{- 5}{1}$

Divide $- 5$ by $1$ .

$d = - 5$

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

$e = 0 - \frac{{(- 810)}^{2}}{4 \cdot 81}$

Simplify the right side.

Simplify each term.

Raise $- 810$ to the power of $2$ .

$e = 0 - \frac{656100}{4 \cdot 81}$

Multiply $4$ by $81$ .

$e = 0 - \frac{656100}{324}$

Divide $656100$ by $324$ .

$e = 0 - 1 \cdot 2025$

Multiply $- 1$ by $2025$ .

$e = 0 - 2025$

Subtract $2025$ from $0$ .

$e = - 2025$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $81 {(x - 5)}^{2} - 2025$ .

$81 {(x - 5)}^{2} - 2025$

Substitute $81 {(x - 5)}^{2} - 2025$ for $81 x^{2} - 810 x$ in the equation $81 x^{2} - y^{2} - 810 x - 10 y = - 2009$ .

$81 {(x - 5)}^{2} - 2025 - y^{2} - 10 y = - 2009$

Move $- 2025$ to the right side of the equation by adding $2025$ to both sides.

$81 {(x - 5)}^{2} - y^{2} - 10 y = - 2009 + 2025$

Complete the square for $- y^{2} - 10 y$ .

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

$a = - 1$

$b = - 10$

$c = 0$

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

$d = \frac{- 10}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10$ .

$d = \frac{2 \cdot - 5}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{- 5}{- 1}$ .

$d = - 1 \cdot - 5$

Rewrite $- 1 \cdot - 5$ as $- - 5$ .

$d = - - 5$

Multiply $- 1$ by $- 5$ .

$d = 5$

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

$e = 0 - \frac{{(- 10)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 0 - \frac{100}{- 4}$

Divide $100$ by $- 4$ .

$e = 0 - - 25$

Multiply $- 1$ by $- 25$ .

$e = 0 + 25$

Add $0$ and $25$ .

$e = 25$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 5)}^{2} + 25$ .

$- {(y + 5)}^{2} + 25$

Substitute $- {(y + 5)}^{2} + 25$ for $- y^{2} - 10 y$ in the equation $81 x^{2} - y^{2} - 810 x - 10 y = - 2009$ .

$81 {(x - 5)}^{2} - {(y + 5)}^{2} + 25 = - 2009 + 2025$

Move $25$ to the right side of the equation by adding $25$ to both sides.

$81 {(x - 5)}^{2} - {(y + 5)}^{2} = - 2009 + 2025 - 25$

Simplify $- 2009 + 2025 - 25$ .

Add $- 2009$ and $2025$ .

$81 {(x - 5)}^{2} - {(y + 5)}^{2} = 16 - 25$

Subtract $25$ from $16$ .

$81 {(x - 5)}^{2} - {(y + 5)}^{2} = - 9$

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

$- 81 {(x - 5)}^{2} + {(y + 5)}^{2} = 9$

Divide each term by $9$ to make the right side equal to one.

$- \frac{81 {(x - 5)}^{2}}{9} + \frac{{(y + 5)}^{2}}{9} = \frac{9}{9}$

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 $1$ .

$\frac{{(y + 5)}^{2}}{9} - \frac{{(x - 5)}^{2}}{\frac{1}{9}} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find the 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, $a$ .

$a = 3$

$b = \frac{1}{3}$

$k = - 5$

$h = 5$

Step 4

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

$y = \pm 9 \cdot (x - (5)) - 5$

Step 5

Simplify to find the first asymptote.

Remove parentheses.

$y = 9 \cdot (x - (5)) - 5$

Simplify $9 \cdot (x - (5)) - 5$ .

Simplify each term.

Multiply $- 1$ by $5$ .

$y = 9 \cdot (x - 5) - 5$

Apply the distributive property.

$y = 9 x + 9 \cdot - 5 - 5$

Multiply $9$ by $- 5$ .

$y = 9 x - 45 - 5$

Subtract $5$ from $- 45$ .

$y = 9 x - 50$

Step 6

Simplify to find the second asymptote.

Remove parentheses.

$y = - 9 \cdot (x - (5)) - 5$

Simplify $- 9 \cdot (x - (5)) - 5$ .

Simplify each term.

Multiply $- 1$ by $5$ .

$y = - 9 \cdot (x - 5) - 5$

Apply the distributive property.

$y = - 9 x - 9 \cdot - 5 - 5$

Multiply $- 9$ by $- 5$ .

$y = - 9 x + 45 - 5$

Subtract $5$ from $45$ .

$y = - 9 x + 40$

Step 7

This hyperbola has two asymptotes.

$y = 9 x - 50, y = - 9 x + 40$

Step 8

The asymptotes are $y = 9 x - 50$ and $y = - 9 x + 40$ .

Asymptotes: $y = 9 x - 50, y = - 9 x + 40$

Step 9

Find the Asymptotes 81x^2-y^2-810x-10y+2009=0

Related Questions

Support LCMGCF.com by disabling your adblocker.