Hello there. To solve this question, we'll have to remember some properties about foci and asymptotes of hyperbolas.
First, the equation for a hyperbola with real axis a and imaginary axis b is:
[tex]\dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{b^2}=1[/tex]
centered at (x0, y0) and axis of symmetry being the x-axis.
The asymptotes are given by the equations:
[tex]y=\pm\dfrac{b}{a}x[/tex]
The relationship between a, b (the real and imaginary axis) and the value c (that stands for the coordinates of the foci)
[tex](-c,0)\text{ and }(c,0)[/tex]
for hyperbolas is:
[tex]c^2=a^2+b^2[/tex]
From the equation of the asymptotes, we find that:
[tex]\begin{gathered} \dfrac{b}{a}x=7x \\ \\ \dfrac{b}{a}=7\Rightarrow b=7a \end{gathered}[/tex]
Hence we get:
[tex]c^2=a^2+(7a)^2=a^2+49a^2=50a^2[/tex]
And from the coordinates of the foci, we get
[tex]c=10[/tex]
Hence
[tex]\begin{gathered} 10^2=50a^2 \\ 100=50a^2 \\ a^2=\frac{100}{50}=2\Rightarrow a=\sqrt{2} \end{gathered}[/tex]
Then from the relation with b, we get
[tex]b=7a=7\sqrt{2}[/tex]
Finally, the center is the midpoint between the foci. In this case, we get:
[tex]\begin{gathered} x_{center}=\dfrac{-10+10}{2}=0 \\ \\ y_{center}=\dfrac{0+0}{2}=0 \end{gathered}[/tex]
So we say the center is at the origin, or (x0, y0) = (0, 0).
Then we find the equation of the hyperbola:
[tex]\dfrac{x^2}{2}-\dfrac{y^2}{98}=1[/tex]
And with a software, you can see this is the right answer (see the following image:)
This is the answer to this question.
