"What happens to the location of the foci of a hyperbola as the value of $a$ becomes increasingly smaller than the value of $b$?"
I assumed that the hyperbola was in the form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
My explanation was as follows:
"As $a \to 0$ then $a^2+b^2 \to b^2$. Because $a^2+b^2=c^2$, this implies that $c^2 \to b^2$ as $a^2$ decreases, and therefore $c \to b$. The foci approach "$b$ away" from the center as the vertices approach the center."
This question was present on the last test I took. I didn't get the question wrong, but when reviewing the test with the class, my teacher said the foci move away from the center. I cannot see how she is right unless we assume $a \to -\infty$. And that is not true because $a > 0$ and $b > 0$ so $c > 0$. My only thought is I misunderstood her.
A last note: I am in a class which is called "Algebra 2 / Trigonometry Honors" at my school, so what was expected was not really suppose to be very formal, but rather just an explanation in words and a diagram (which I provided).
Edit: She could be right if $b \to \infty$ instead of $a \to 0$. But I think my interpretation of the question makes more sense given the question. If someone cares to share their opinion please do. I am just trying to figure out what the question actually asks and a more formal way to justify and prove my answer. Thanks!
Without loss of generality, take the origin to be the center: $$ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1. $$ The foci are $\pm(\sqrt{a^{2} + b^{2}}, 0)$, and the question, taken literally, reads, "What happens to the foci as $a/b \to 0$?"
Strictly speaking, that's a dicey question: In the absence of additional constraints, $a$ and $b$ are independent, so "there are lots of ways to take $a/b \to 0$". (Geometrically, the absolute slope of an asymptote and the overall scale are independent.)
To approach matters systematically, write $$ c = \sqrt{a^{2} + b^{2}} = b\sqrt{(a/b)^{2} + 1}. $$ As $a/b \to 0$, clearly $c/b \to 1$. Geometrically, we might say the foci approach $\pm b$ compared to the overall scale of the hyperbola.
In more detail: If $b$ approaches a finite limit as $a/b \to 0$, then $c \to b$, just as you said on your test.
In general, the trick of "multiplying and dividing by the conjugate expression" gives $$ c - b = b\bigl[\sqrt{(a/b)^{2} + 1} - 1\bigr] = \frac{a^{2}/b}{\sqrt{(a/b)^{2} + 1} + 1}. $$ From here, it's easy to see $c \to b$ can fail.
If $a = t$ and $b = t^{3/2}$, for example, then as $t \to \infty$ we have $a/b = 1/\sqrt{t} \to 0$, but $$ c - b = \frac{\sqrt{t}}{\sqrt{(1/t) + 1} + 1} \to \infty. $$ If instead $k > 0$ is real, $a = t$, and $b = t^{2}/(2k)$, then $a/b = 2k/t \to 0$ but $$ c - b = \frac{2k}{\sqrt{(2k/t)^{2} + 1} + 1} \to k. $$ That is, the asymptotic distance from $b$ to $c$ can be any non-negative real number, or infinity. (In fact, $c - b$ can exhibit arbitrarily complicated behavior as $a/b \to 0$: Just replace the constant $k$ by your favorite bounded, non-vanishing function of $t$.)
Whether or not the foci get further from the center depends on how $a/b \to 0$. Your teacher presumably made some assumption, such as "$a$ is constant and $b$ increases without bound".
Overall, it appears the question was more complicated than the questioner intended.