Question

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

Answer 1

Answer:  8

Step-by-step explanation:

x² - 2x  + ___ - 4y²  = 15 + ___

x² - 2x  + 1 - 4y²  = 15 + 1

(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