Question

HELLLLP!!!!

Answer 1

Answer:

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

Answer 2

Answer:

$w=8\\h=4$

Step-by-step 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.

$x^{2} -2x-4y^{2}=15\\x^{2} -2x+(\frac{2}{2} )^{2} -4y^{2} =15+1\\ (x-1)^{2}-4y^{2}=16\\\frac{(x-1)^{2} }{16} -\frac{4y^{2} }{16} =\frac{16}{16}\\\frac{(x-1)^{2} }{16} -\frac{y^{2} }{4}=1$

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.

$2(4)=8\\2(2)=4$