Answer:
The co-vertices are (1 , 4) , (1 , 0)
Explanation:
* Lets study the equation of the hyperbola
- The standard form of the equation of a hyperbola with
center (h , k) and transverse axis parallel to the x-axis is
(x - h)²/a² - (y - k)²/b² = 1 ,Where
# The length of the transverse axis is 2a
# The coordinates of the vertices are (h ± a , k)
# The length of the conjugate axis is 2b
# The coordinates of the co-vertices are (h , k ± b)
# The coordinates of the foci are (h ± c , k), where c² = a² + b²
* Lets solve the problem
- From the graph
∵ The vertices of the hyperbola are (6 , 2) and (-4 , 2)
∵ The length of the transverse axis is 2a
- The length of the transverse axis is the difference
between the x-coordinates of the vertices
∴ The length of the transverse axis = 6 - (-4) = 6 + 4 = 10
∴ 2a = 10 ⇒ divide both sides by 2
∴ a = 5
∵ The coordinates of the vertices are (h + a , k) and (h - a , k)
∵ The vertices of the hyperbola are (6 , 2) and (-4 , 2)
∴ k = 2
∴ h + a = 6 and h - a = -4
∵ a = 5
∴ h + 5 = 6 ⇒ subtract 5 from both sides
∴ h = 1
∴ The center of the hyperbola is (1 , 2)
∵ The length of the conjugate axis is 2b
- The length of the conjugate axis is the difference
between the y-coordinates of the co-vertices
∴ The length of the conjugate axis = 4 - 0 = 4
∴ 2b = 4 ⇒ divide both sides by 2
∴ b = 2
∵ The coordinates of the co-vertices are (h , k + b) and (h , k - b)
∵ h = 1 , k = 2 , b = 2
∴ The co-vertices are (1 , 2 + 2) and (1 , 2 - 2)
∴ The co-vertices are (1 , 4) , (1 , 0)