Respuesta :
Step-by-step explanation:
Here any equation of hyperbola is given as:
x^2/a^2 - y^2/b^2 = 1
And as its x intercept is 3, that means its one vertex is (3,0) that is the value of (a,0).
So, a = 3
y= 2(3) = 6
So, on plug in these values un above equation,
x^2/3^2 - y^2/6^2 = 1
or
x^2/9 - y^2/36 = 1
THAT'S THE REQUIRED EQUATION OF HYPERBOLA!!!
Hope this helps :)
Answer:
[tex]\large \boxed{\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 centre 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. Calculate a
2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6
a = 6/2 = 3
3. Calculate b
The equation for the asymptote with the positive slope is
[tex]y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x[/tex]
So, the slope of the asymptote is
[tex]\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}[/tex]
4. Write the equation of the hyperbola
The equation of the hyperbola is
[tex]\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}[/tex]
The figure below shows your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.
