Answer
See graphs and explanations below.
Work Step by Step
(a) Given the hyperbola as $\frac{y^2}{k}-\frac{x^2}{16-k} =1$, we have $a^2=k, b^2=16-k$. Use the formula $c^2=a^2+b^2$, we have $c^2=k+16-k=16$ and $c=4$. Thus the hyperbolas are confocal with foci at $(0,\pm4)$
(b) See graphs, as $k$ increases, the absolute value of the asymptote slopes increases also. as a result, the opening of the curves becomes narrower.
