The equation of a hyperbola when the transverse axis is horizontal is given as
[tex]\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{where c}^2=a^2+b^2 \end{gathered}[/tex]
Given that a=6, c=8, let us solve for b using
[tex]\begin{gathered} c^2=a^2+b^2 \\ b^2=c^2-a^2 \\ b^2=8^2-6^2^{} \\ b^2=64-36 \\ b^2=28 \end{gathered}[/tex]
Note that b²=28 and a²=6²=36
Substituting into the equation will give
[tex]\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \frac{x^2}{36}-\frac{y^2}{28}=1 \end{gathered}[/tex]
Hence, the correct answer is the first option
