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

User Bdrelling
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1 Answer

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Answer: 8

Explanation:

x² - 2x + ___ - 4y² = 15 + ___

x² - 2x + 1 - 4y² = 15 + 1

(x - 1)² - 4(y)² = 16


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


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


r_(x) = 4
r_(y) = 2

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

User Bojangles
by
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