Respuesta :
The given vertices are (-9,0) and (9,0).
Notice that they lie on the x-axis since they have 0 as their y-coordinate.
Hence, the hyperbola is a horizontal hyperbola.
Recall that the equation of a horizontal hyperbola is given as:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]Where (h,k) is the center and a>b.
As both vertices are equidistant from the origin, the center of the hyperbola is (0,0), and the equation becomes:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]Note that the vertices are at (-a,0) and (a,0).
Compare with the given vertices (-9,0) and (9,0). It follows that a=9.
Substitute this into the equation:
[tex]\frac{x^2}{9^2}-\frac{y^2}{b^2}=1[/tex]Recall that the length of the conjugate axis is given as 2b, it follows that:
[tex]\begin{gathered} 2b=16 \\ \Rightarrow b=\frac{16}{2}=8 \end{gathered}[/tex]Substitute b=8 into the equation:
[tex]\begin{gathered} \frac{x^2}{9^2}-\frac{y^2}{8^2}=1 \\ \Rightarrow\frac{x^2}{81}-\frac{y^2}{64}=1 \end{gathered}[/tex]The required equation in standard form is:
[tex]\frac{x^2}{81}-\frac{y^2}{64}=1[/tex]