Question

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

Answer 1

`-(x-1)^2/(sqrt15)^2+(y-10)^2/(sqrt10)^2=1`

Explanation:

`-2x^2+3y^2+4x-60y+268=0`
Rearranging
`-2(x^2-2x)+3(y^2-20y)=-268`
Completing the squares
`-2(x^2-2x+1)+3(y^2-20y+100)=-268-2+300`
`-2(x-1)^2+3(y-10)^2=30`
Dividing by 30
`-(x-1)^2/15+(y-10)^2/10=1`
In standard form
`-(x-1)^2/(sqrt15)^2+(y-10)^2/(sqrt10)^2=1`

Answer 2

Please see the explanation for the process.

`(y - 10)^2/(sqrt(10))^2 -(x - 1)^2/(sqrt(15))^2 = 1`

Explanation:

Add `3k^2 - 2h^2 - 268` to both sides:

`-2x^2 + 4x - 2h^2 + 3y^2 - 60y + 3k^2 = 3k^2 - 2h^2 - 268`

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

`-2(x^2 - 2x + h^2) + 3(y^2 - 20y + k^2) = 3k^2 - 2h^2 - 268`

Using the pattern, `(x - h)^2 = x^2 -2hx + h^2`, set the right side of the pattern equal to the first three terms:

`x^2 -2hx + h^2 = x^2 - 2x + h^2`

Combine like terms:

`-2hx = - 2x`

Solve for `h` and compute `h^2`:

`h = 1, h^2 = 1`

Substitute `(x - 1)^2` for `x^2 - 2x + h^2` and `-2` for `-2h^2`

`-2(x - 1)^2 + 3(y^2 - 20y + k^2) = 3k^2 - 2 - 268`

Using the pattern, `(y - k)^2 = y^2 -2ky + k^2`, set the right side of the pattern equal to the second three terms:

`y^2 -2ky + k^2 = y^2 - 20y + k^2`

Combine like terms:

`-2ky = -20y`

`k = 10, k^2 = 100`

Substitute `(y - 10)^2` for `y^2 - 20y + k^2` and `300` for `3k^2`

`-2(x - 1)^2 + 3(y - 10)^2 = 300 - 2 - 268`

Combine the terms on the right:

`-2(x - 1)^2 + 3(y - 10)^2 = 30`

Divide both sides by 30:

`(y - 10)^2/10 -(x - 1)^2/15 = 1`

Write the denominators as squares:

`(y - 10)^2/(sqrt(10))^2 -(x - 1)^2/(sqrt(15))^2 = 1`