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

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step-by-step explanation:

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