Answer
(a) See graph and explanations.
(b) same asymptotes.
(c) See explanations.
Work Step by Step
(a) Divide 16 for both equations and rewrite them in standard forms as $\frac{x^2}{4^2} -\frac{y^2}{2^2}=-1 $ and $\frac{x^2}{4^2} -\frac{y^2}{2^2}=1 $. Compare these two equation with the definition of the conjugate hyperbolas given at the beginning of this Exercise, we conclude that the two hyperbolas are conjugate to each other. See graph for the two hyperbolas.
(b) Based on the observations, these two hyperbolas have the same asymptotes.
(c) We start with the universal conjugate hyperbola equations given at the beginning of the Exercise. The first hyperbola has asymptotes as $y=\pm\frac{b}{a}x$. Rewrite the second equation as $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1 $, this hyperbola has vertical transverse axis of length $2b$, and the asymptotes are $y=\pm\frac{b}{a}x$. Thus the two conjugate hyperbolas have the same asymptotes.
