Answer:
[tex]{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}[/tex]
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 [tex]\dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1[/tex]
The center is at (h,k).
The distance between the vertices is 2a.
The equations of the asymptotes are[tex]y = k \pm \dfrac{b}{a}(x - h)[/tex]
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
[tex]y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x[/tex]
Thus, asymptote has the slope of
[tex]\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}[/tex]
4. The equation of the hyperbola is
[tex]\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}[/tex]
The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.
