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What is the equation of a hyperbola centered at the origin with x-intercepts ±3 and asymptote y=2x?

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 :)

znk

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.

Ver imagen znk