Answer:
1) (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1
2) (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1
3) (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1
4) (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1
Step-by-step 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
- the coordinates of the foci are (h ± c , k), where c² = a² + b²
- 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 coordinates of the foci are (h , k ± c), where c² = a² + b²
* Lets look to the problem
1) The foci are (1 , -22) and (1 , 12)
- Compare the point with the previous rules
∵ h = 1 and k ± c = -22 ,12
∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1
∵ k + c = -22 ⇒ (1)
∵ k - c = 12 ⇒ (2)
* Add (1) and(2)
∴ 2k = -10 ⇒ ÷2
∴ k = -5
* substitute the value of k in (1)
∴ -5 + c = -22 ⇒ add 5 to both sides
∴ c = -17
∴ c² = (-17)² = 289
∵ c² = a² + b²
∴ a² + b² = 289
* Now lets check which answer has (h , k) = (1 , -5)
and a² + b² = 289 in the form (y - k)²/a² - (x - h)²/b² = 1
∵ 15² + 8² = 289
∵ (h , k) = (1 , -5)
∴ The answer is (y + 5)²/15² - (x - 1)²/8² = 1
* (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1
2) The foci are (-7 , 5) and (3 , 5)
- Compare the point with the previous rules
∵ k = 5 and h ± c = -7 ,3
∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1
∵ h + c = -7 ⇒ (1)
∵ h - c = 3 ⇒ (2)
* Add (1) and(2)
∴ 2h = -4 ⇒ ÷2
∴ h = -2
* substitute the value of h in (1)
∴ -2 + c = -7 ⇒ add 2 to both sides
∴ c = -5
∴ c² = (-5)² = 25
∵ c² = a² + b²
∴ a² + b² = 25
* Now lets check which answer has (h , k) = (-2 , 5)
and a² + b² = 25 in the form (x - h)²/a² - (y - k)²/b² = 1
∵ 3² + 4² = 25
∵ (h , k) = (-2 , 5)
∴ The answer is (x + 2)²/3² - (y - 5)²/4² = 1
* (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1
3) The foci are (-6 , -2) and (14 , -2)
- Compare the point with the previous rules
∵ k = -2 and h ± c = -6 ,14
∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1
∵ h + c = -6 ⇒ (1)
∵ h - c = 14 ⇒ (2)
* Add (1) and(2)
∴ 2h = 8 ⇒ ÷2
∴ h = 4
* substitute the value of h in (1)
∴ 4 + c = -6 ⇒ subtract 4 from both sides
∴ c = -10
∴ c² = (-10)² = 100
∵ c² = a² + b²
∴ a² + b² = 100
* Now lets check which answer has (h , k) = (4 , -2)
and a² + b² = 100 in the form (x - h)²/a² - (y - k)²/b² = 1
∵ 8² + 6² = 100
∵ (h , k) = (4 , -2)
∴ The answer is (x - 4)²/8² - (y + 2)²/6² = 1
* (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1
4) The foci are (-7 , -10) and (-7 , 16)
- Compare the point with the previous rules
∵ h = -7 and k ± c = -10 , 16
∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1
∵ k + c = -10 ⇒ (1)
∵ k - c = 16 ⇒ (2)
* Add (1) and(2)
∴ 2k = 6 ⇒ ÷2
∴ k = 3
* substitute the value of k in (1)
∴ 3 + c = -10 ⇒ subtract 3 from both sides
∴ c = -13
∴ c² = (-13)² = 169
∵ c² = a² + b²
∴ a² + b² = 169
* Now lets check which answer has (h , k) = (-7 , 3)
and a² + b² = 169 in the form (y - k)²/a² - (x - h)²/b² = 1
∵ 5² + 12² = 169
∵ (h , k) = (-7 , 3)
∴ The answer is (y - 3)²/5² - (x + 7)²/12² = 1
* (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1