# How do you write the standard form of the hyperbola -2x^2+3y^2+4x-60y+268=0?

Oct 17, 2016

$- {\left(x - 1\right)}^{2} / {\left(\sqrt{15}\right)}^{2} + {\left(y - 10\right)}^{2} / {\left(\sqrt{10}\right)}^{2} = 1$

#### Explanation:

$- 2 {x}^{2} + 3 {y}^{2} + 4 x - 60 y + 268 = 0$
Rearranging
$- 2 \left({x}^{2} - 2 x\right) + 3 \left({y}^{2} - 20 y\right) = - 268$
Completing the squares
$- 2 \left({x}^{2} - 2 x + 1\right) + 3 \left({y}^{2} - 20 y + 100\right) = - 268 - 2 + 300$
$- 2 {\left(x - 1\right)}^{2} + 3 {\left(y - 10\right)}^{2} = 30$
Dividing by 30
$- {\left(x - 1\right)}^{2} / 15 + {\left(y - 10\right)}^{2} / 10 = 1$
In standard form
$- {\left(x - 1\right)}^{2} / {\left(\sqrt{15}\right)}^{2} + {\left(y - 10\right)}^{2} / {\left(\sqrt{10}\right)}^{2} = 1$

Oct 17, 2016

Please see the explanation for the process.

${\left(y - 10\right)}^{2} / {\left(\sqrt{10}\right)}^{2} - {\left(x - 1\right)}^{2} / {\left(\sqrt{15}\right)}^{2} = 1$

#### Explanation:

Add $3 {k}^{2} - 2 {h}^{2} - 268$ to both sides:

$- 2 {x}^{2} + 4 x - 2 {h}^{2} + 3 {y}^{2} - 60 y + 3 {k}^{2} = 3 {k}^{2} - 2 {h}^{2} - 268$

Remove a factor of -2 from the first 3 terms and a factor of 3 from the next 3 terms:

$- 2 \left({x}^{2} - 2 x + {h}^{2}\right) + 3 \left({y}^{2} - 20 y + {k}^{2}\right) = 3 {k}^{2} - 2 {h}^{2} - 268$

Using the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, set the right side of the pattern equal to the first three terms:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} - 2 x + {h}^{2}$

Combine like terms:

$- 2 h x = - 2 x$

Solve for $h$ and compute ${h}^{2}$:

$h = 1 , {h}^{2} = 1$

Substitute ${\left(x - 1\right)}^{2}$ for ${x}^{2} - 2 x + {h}^{2}$ and $- 2$ for $- 2 {h}^{2}$

$- 2 {\left(x - 1\right)}^{2} + 3 \left({y}^{2} - 20 y + {k}^{2}\right) = 3 {k}^{2} - 2 - 268$

Using the pattern, ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$, set the right side of the pattern equal to the second three terms:

${y}^{2} - 2 k y + {k}^{2} = {y}^{2} - 20 y + {k}^{2}$

Combine like terms:

$- 2 k y = - 20 y$

$k = 10 , {k}^{2} = 100$

Substitute ${\left(y - 10\right)}^{2}$ for ${y}^{2} - 20 y + {k}^{2}$ and $300$ for $3 {k}^{2}$

$- 2 {\left(x - 1\right)}^{2} + 3 {\left(y - 10\right)}^{2} = 300 - 2 - 268$

Combine the terms on the right:

$- 2 {\left(x - 1\right)}^{2} + 3 {\left(y - 10\right)}^{2} = 30$

Divide both sides by 30:

${\left(y - 10\right)}^{2} / 10 - {\left(x - 1\right)}^{2} / 15 = 1$

Write the denominators as squares:

${\left(y - 10\right)}^{2} / {\left(\sqrt{10}\right)}^{2} - {\left(x - 1\right)}^{2} / {\left(\sqrt{15}\right)}^{2} = 1$