Write the equation of a hyperbola with vertices (3, - 1) and (3, - 9) and co-vertices (-6. - 5) and (12, - 5).

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The equation of the hyperbola is (y + 5) ²/16 - (x - 3) ²/81 = 1

Step-by-step explanation:

* Lets revise the equation of the hyperbola

* The standard form of the equation of a hyperbola with

center (h, k) and transverse axis parallel to the y-axis is

(y - k) ²/a² - (x - h) ²/b² = 1

- The length of the transverse axis is 2a

- The coordinates of the vertices are (h, k ± a)

- The length of the conjugate axis is 2b

- The coordinates of the co-vertices are (h ± b, k)

- The distance between the foci is 2c, where c² = a² + b²

- The coordinates of the foci are (h, k ± c)

* Lets solve the problem

∵ The vertices of the hyperbola are (3, - 1), (3, 9)

∵ The coordinates of its vertices are (h, k + a) and (h, k - a)

∴ h = 3

∴ k + a = - 1 and k - a = - 9

∵ The co-vertices of it are (-6, - 5) and (12, - 5)

∵ The vertices of the co-vertices are (h + b, k) and (h - b, k)

∴ k = - 5

∴ h + b = - 6 and h - b = 12

∵ h = 3

∴ 3 + b = - 6 ⇒ subtract 3 from both sides

∴ b = - 9

∵ k + a = - 1

∵ k = - 5

∴ - 5 + a = - 1 ⇒ add 5 to both sides

∴ a = 4

∵ The equation of the hyperbola is (y - k) ²/a² - (x - h) ²/b² = 1

∵ a = 4, b = - 9, h = 3, k = - 5

∴ The equation of the hyperbola is (y - - 5) ² / (4) ² - (x - 3) ² / (-9) ² = 1

∴ The equation of the hyperbola is (y + 5) ²/16 - (x - 3) ²/81 = 1