Question

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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.

Answer 1

`w=8\\h=4`

Explanation:

The given equation is

`x^(2) -4y^(2)-2x-15=0`

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

`x^(2) -2x-4y^(2)=15\\x^(2) -2x+((2)/(2) )^(2) -4y^(2) =15+1\\ (x-1)^(2)-4y^(2)=16\\((x-1)^(2) )/(16) -(4y^(2) )/(16) =(16)/(16)\\((x-1)^(2) )/(16) -(y^(2) )/(4)=1`

Now that we have the explicit form, you can observe that
`a^(2)=16 \implies a=4` and
`b^(2)=4 \implies b=2`.

On the other hand, the width of the asymptote rectangle is
`2a` and the height is
`2b`.

Therefore, the dimensions are 8 by 4.

`2(4)=8\\2(2)=4`

Answer 2

The width the asymptote rectangle is 8 units

The height the asymptote rectangle is 4 units

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