Question

Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x

Answer 1

Answer:

${\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}$

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  $\dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1$

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes are$y = k \pm \dfrac{b}{a}(x - h)$

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

$y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x$

Thus,  asymptote has the slope of

$\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}$

4.  The equation of the hyperbola is

$\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}$

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.