Question

Which is the corresponding equation of a hyperbola with vertices at (11, 0) and (-11, 0) and co-vertices (0, 22) and (0, -22).

Answer 1

$B)\frac{x^2}{121}-\frac{y^2}{484}=1$

1) Considering that the vertices and co-vertices are: (11,0) and (-11,0) and (0,22) and (0,-22) we can visualize that this hyperbola has their arches passing through the x-axis.

2) We can then, write out the following to get the equation of the hyperbola:

$\begin{gathered} \frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1 \\ h=0,k=0 \\ a^2=\left(k+11\right)^2\Rightarrow a=11 \\ b^2=\left(k-22\right)^2\operatorname{\Rightarrow}b=22 \\ \\ \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \\ \\ \frac{(x-0)^2}{11^2}-\frac{(y-0)^2}{22^2}=1 \\ \\ \frac{x^2}{121}-\frac{y^2}{484}=1 \end{gathered}$

Note that this is the vertex form. So we can tell the answer is:

$\frac{x^2}{121}-\frac{y^2}{484}=1$