Question

Use the information provided to write the standard form equation of each hyperbola.

Answer 1

9x^2-4y^2-180x+64y+320

Answer 2

There should be a "= 0" at the end.
9x^2 - 4y^2 - 180x + 64y + 320 = 0

Take the constant to the other side:
9x^2 - 4y^2 - 180x + 64y = -320

Group the x terms together. Group the y terms together:
9x^2 - 180x -4y^2 + 64y = -320
9(x^2 - 20x) -4(y^2 - 16y) = -320

Complete the squares:
9{ (x^2 - 20x + 100) - 100 } -4{ (y^2 - 16y + 64) - 64 } = -320
9{ (x - 10)^2 - 100 } -4{ (y - 8)^2 - 64 } = -320
9(x - 10)^2 - 900 - 4(y - 8)^2 + 256 = -320
9(x - 10)^2 -4(y - 8)^2 = -320 + 900 - 256
9(x - 10)^2 -4(y - 8)^2 = 324

Divide by 324:
9/324(x - 10)^2 -4/324(y - 8)^2 = 1
1/36(x - 10)^2 - 1/81(y - 8)^2 = 1
(x - 10)^2 / 6^2 - (y - 8)^2 / 9^2 = 1

And that is the standard form for a hyperbola.

Answer 3

\[\frac{ (x - 10)^{2} }{ 6^{2} } - \frac{ (y - 8)^{2} }{ 9^{2}} = 1\]