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HELLLLP!!!!
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.

HELLLLP 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 class=

Respuesta :

Ashraf82

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:

[tex]w=8\\h=4[/tex]

Step-by-step explanation:

The given equation is

[tex]x^{2} -4y^{2}-2x-15=0[/tex]

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

[tex]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[/tex]

Now that we have the explicit form, you can observe that [tex]a^{2}=16 \implies a=4[/tex] and [tex]b^{2}=4 \implies b=2[/tex].

On the other hand, the width of the asymptote rectangle is [tex]2a[/tex] and the height is [tex]2b[/tex].

Therefore, the dimensions are 8 by 4.

[tex]2(4)=8\\2(2)=4[/tex]

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