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charle [14.2K]
3 years ago
11

The equation of a hyperbola is x^2-4y^2-2x-15=0. What is the width of the asymptote rectangle?

Mathematics
1 answer:
zhenek [66]3 years ago
3 0

Answer:  8

<u>Step-by-step explanation:</u>

x² - 2x  + ___ - 4y²  = 15 + ___

x² - 2x  + <u> 1 </u> - 4y²  = 15 + <u> 1 </u>

(x - 1)² - 4(y)² = 16

\frac{(x-1)^{2}} {16} - \frac{4(y - 0)^{2}}{16} = \frac{16}{16}

\frac{(x-1)^{2}} {16} - \frac{(y - 0)^{2}}{4} = 1

r_{x} = 4   r_{y} = 2

width of rectangle is 2(r_{x}) = 2(4) = 8

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