type the correct answer in each box.
the equation of a hyperbola is x2 − 4y2 − 2x − 15 = 0.
the width the asymptote rectangle is units, and its height is units.

The width the asymptote rectangle is 8 units

The height the asymptote rectangle is 4 units

Step-by-step 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

- The length of the transverse axis is 2a (the width of the rectangle)

- The length of the conjugate axis is 2b (the height of the rectangle)

- 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 - 15 = 0

∴ (x² - 2x) + (-4y²) - 15 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y²) - 15 = 0

∴ (x² - 2x) - 4(y²) - 15 = 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 = (x - 1)² - 1

- Lets put the equation after making the completing square

∴ (x - 1)² - 1 - 4(y²) - 15 = 0 ⇒ simplify

∴ (x - 1)² - 4(y)² - 16 = 0 ⇒ add the two side by 16

∴ (x - 1)² - 4(y)² = 16 ⇒ divide both sides by 16

∴ (x - 1)²/16 - y²/4 = 1

∴ (x - 1)²/16 - y²/4 = 1

∴ The standard form of the equation of the hyperbola is

  (x - 1)²/16 - y²/4 = 1

∵ The standard form of the equation of hyperbola is

  (x - h)²/a² - (y - k)²/b² = 1

∴ a² = 16 and b² = 4

∴ a = 4 , b = 2

∵ The width the asymptote rectangle is 2a

∴ The width the asymptote rectangle = 2 × 4 = 8 units

∵ The height the asymptote rectangle is 2b

∴ The height the asymptote rectangle = 2 × 2 = 4 units

Step-by-step explanation:

The given equation is

First, we complete squares for each variable to find the explicit form of the hyperbola.

Now that we have the explicit form, you can observe that and .

On the other hand, the width of the asymptote rectangle is and the height is .

Therefore, the dimensions are 8 by 4.

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step-by-step explanation:

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