Create the equation of a hyperbola centered at the origin, with a horizontal transverse axis, vertex at (–7, 0), and asymptotes of y equals plus or minus six sevenths x period Show your work. (4 points)

Create the equation of a hyperbola centered at the origin with a horizontal transverse axis vertex at 7 0 and asymptotes of y equals plus or minus six sevenths class=

Respuesta :

So,

First, remember that the general equation of a hyperbola is given by the following:

[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]

In our problem, we're given that this hyperbola is centered at the origin, so C(h,k) = C(0,0). Then, h=0, and k=0.

Our equation can be written as:

[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]

The vertex of this hyperbola is set in the point (-7,0). We could use the following equation to find the value of a:

[tex]V=(h\pm a,k)[/tex]

Where V is the vertex.

If we replace our values:

[tex](0\pm a,0)=(-7,0)[/tex]

So,

[tex]a=\pm7[/tex]

Since "a" is squared in the equation, it doesn't matter if it is -7 or 7. The +/- sign indicates that there are two vertices at (-7,0) and (7,0) respectively.

Now, to find b, we could use the fact that the asymptotes of the hyperbola are:

[tex]y=\pm\frac{6}{7}[/tex]

Remember that these asymptotes have the form:

[tex]y=\pm\frac{b}{a}[/tex]

So, if we compare:

[tex]\begin{gathered} a=\pm7 \\ b=\pm6 \end{gathered}[/tex]

And, our equation will be:

[tex]\frac{x^2}{49}-\frac{y^2}{36}=1[/tex]

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