Answer:
(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices
Explanation:
* Lets explain how to solve this problem
- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0
- The standard form of the equation of hyperbola is
(x - h)²/a² - (y - k)²/b² = 1 where a > b
- So lets collect x in a bracket and make it a completing square and
also collect y in a bracket and make it a completing square
∵ x² - 4y² - 2x + 16y - 31 = 0
∴ (x² - 2x) + (-4y² + 16y) - 31 = 0
- Take from the second bracket -4 as a common factor
∴ (x² - 2x) + -4(y² - 4y) - 31 = 0
∴ (x² - 2x) - 4(y² - 4y) - 31 = 0
- Lets make (x² - 2x) completing square
∵ √x² = x
∴ The 1st term in the bracket is x
∵ 2x ÷ 2 = x
∴ The product of the 1st term and the 2nd term is x
∵ The 1st term is x
∴ the second term = x ÷ x = 1
∴ The bracket is (x - 1)²
∵ (x - 1)² = (x² - 2x + 1)
∴ To complete the square add 1 to the bracket and subtract 1 out
the bracket to keep the equation as it
∴ (x² - 2x + 1) - 1
- We will do the same withe bracket of y
- Lets make 4(y² - 4y) completing square
∵ √y² = y
∴ The 1st term in the bracket is x
∵ 4y ÷ 2 = 2y
∴ The product of the 1st term and the 2nd term is 2y
∵ The 1st term is y
∴ the second term = 2y ÷ y = 2
∴ The bracket is 4(y - 2)²
∵ 4(y - 2)² = 4(y² - 4y + 4)
∴ To complete the square add 4 to the bracket and subtract 4 out
the bracket to keep the equation as it
∴ 4[y² - 4y + 4) - 4]
- Lets put the equation after making the completing square
∴ (x - 1)² - 1 - 4[(y - 2)² - 4] - 31 = 0 ⇒ simplify
∴ (x - 1)² - 1 - 4(y - 2)² + 16 - 31 = 0 ⇒ add the numerical terms
∴ (x - 1)² - 4(y - 2)² - 16 = 0 ⇒ add 14 to both sides
∴ (x - 1)² - 4(y - 2)² = 16 ⇒ divide both sides by 16
∴ (x - 1)²/16 - (y - 2)²/4 = 1
∵ 16 = (4)² and 4 = (2)²
∴ The standard form of the equation of the hyperbola is
(x - 1)²/4² - (y - 2)²/2² = 1