Disclaimer: The answer below responded to the original version of the OP's question, which was quite different in nature and less specific than the current version.
$p(x∣w_1)$ and $p(x∣w2)$ are equal.
OK, this is going to take a lot longer to answer.
In some statistical applications, a statistician (or a machine, since
you included machine learning as a tag) needs to decide which of two
hypotheses is true: $H_1 \colon w = w_1$ and $H_2 \colon w = w_2$.
It is known that $$P(w = w_1) = P(w = w_2) = \frac{1}{2}.$$
This is what the equal a priori probabilities that you keep referring
to means.
Here is a simple method: Always decide that $w = w_1$, and so hypothesis
$H_1$ is always the true hypothesis. When in fact $H_1$ is true, your
decision is
perfectly correct; when in fact $H_2$ is true your decision is
perfectly wrong, and
thus you have a $50\%$ chance of making an error. More sophisticated methods
use a coin toss or a call to a random number generator to decide, but unfortunately still have a $50\%$ chance of making an an error; the same
as the simpler mulish insistence that $H_1$ is always true.
To get better performance, i.e., smaller error probability), the
statistician might observe a random variable whose distribution depends
on the value of $w$. If $w = w_1$, the distribution is $p(x\mid w_1)$;
if $w = w_2$, the distribution is $p(x\mid w_2)$. For example, if
$w = w_1$, $x$ is a normal random variable with mean $100$ and variance $1$,
while if $w = w_2$, $x$ is a standard normal random variable with mean $0$ and
variance $1$. So if the statistician observes that $x$ has value $101.2$,
it is highly likely that $w = w_1$ and thus very likely that
$H_1$ is true because a standard
normal random variable is quite unlikely to have large value. On the other
hand, if $x$ has small value (say between $-4$ and $+4$), then it is quite
likely that $H_2$ is true and $w = w_2$. But notice that all this depends
critically on the distributions $p(x\mid w_1)$ and $p(x\mid w_2)$ being
different. If the distributions are the same, then observing $x$ is of
no help in deciding between $H_1$ and $H_2$. Thus when you claim that
$p(x∣w_1)$ and $p(x∣w_2)$ are equal
you are effectively insisting that observing $x$ is useless as
far as deciding between $H_1$ and $H_2$ is concerned.
So, how are these distributions known in the first place? The client might provide
them to the statistician based on the knowledge of how the client's
apparatus works. Your professor, like Professor Indiana Jones in the
movie Raiders of the Lost Ark, might
be making them up as he goes along (Remember that $99\frac{44}{100}\%$
of all statistics are made up!). In the context of machine
learning, there may be training samples provided: Here are
$200$ observations of $x$ when $H_1$ is true, and here are
$200$ more when $H_2$ is true. (In your particular problem,
$x$ is a bivariate normal random variable with independent
(standard normal) components when $H_1$ is true and correlated
normal components when
$H_2$ is true, and so each sample would be a a pair of numbers).
The machine estimates
$p(x\mid w_1)$ from the first set of observations
and $p(x\mid w_2)$ from the second set, and uses these
estimates when making decisions when the real work comes
along.
In summary, your claim that $p(x\mid w_1) = p(x\mid w_2)$
means that $x$ is totally useless in distinguishing the two
cases. For your particular distribution, equality holds
(if you nevertheless contiunue to insist on equality)
exactly when $a=b=1$ and $c=d=e=0$ (in which case $ab-c^2 = 1$
as desired). There is no way of solving for $a,b,c,d,e$,
or saying what values of $a,b,c,d,e$ make sense in your
problem based on the information that you have provided.
You need to be given these by your professor,
or you need to be given training data so that you can estimate
these parameters, or you should emulate Professor Jones
and make up some numbers (subject to the constraints that $ab - c^2 = 1$
and $a, b > 0$) and solve the problem using these.
I will use $(X,Y)$ for the observation. Given $w = w_1$, we have that the variances of $X$ and $Y$ are $a$ and $b$ respectively, while the covariance is $c$.
Thus, the correlation coefficient $\rho = \frac{c}{\sqrt{ab}}$ and so
$1-\rho^2 = 1 - \frac{c^2}{ab} = \frac{1}{ab}$. The two conditional (joint) densities are joint normal densities given by
$$\begin{align}
f_1(x,y) &= \frac{1}{2\pi \sqrt{ab(1-\rho^2)}}\exp\left[-\frac{1}{2(1-\rho^2)}\left(\frac{x^2}{a} - 2\rho\frac{xy}{\sqrt{ab}} + \frac{y^2}{b}\right)\right]\\
&= \frac{1}{2\pi}\exp\left[-\frac{ab}{2}\left(\frac{x^2}{a} - 2\frac{c}{\sqrt{ab}}\frac{xy}{\sqrt{ab}} + \frac{y^2}{b}\right)\right]\\
&= \frac{1}{2\pi}\exp\left[-\frac{1}{2}\left(bx^2 - 2cxy + ay^2\right)\right],\\
f_2(x,y)
&= \frac{1}{2\pi}\exp\left[-\frac{1}{2}\left((x-d)^2+(y-e)^2\right)\right]\\
\end{align}$$
The decision boundary is the set of all $(x,y)$ for which $f_1(x,y) = f_2(x,y)$, and so the decision boundary is the conic section specified by
$$bx^2 -2cxy +ay^2 - \left((x-d)^2 + (y-e)^2\right) = 0.$$
Edit: As pointed out in the comment by whuber,
this conic section can be a hyperbola as well as an ellipse or
parabola (including as a special case a straight line).
The discriminant $c^2-(a-1)(b-1) = (a+b)-2$ (since we are
given that $ab-c^2=1$)
can be positive, negative, or zero depending on the variances
$a$ and $b$.
I suspect that this will be work out to be
either an ellipse or a parabola (but not
a hyperbola) depending on
the parameters $a,b,c,d,e$, including, as a special case of parabola,
a straight line. A specific example of a straight-line decision
boundary is when $c = 0$ and $a = b = 1$ so that the
only difference between the two conditional distributions is the means:
$X$ and $Y$ are conditionally independent unit-variance
normal random variables under either hypothesis. In this instance,
the decision boundary is the perpendicular bisector of the
straight line segment with end-points $(0,0)$ and $(d,e)$.
What puzzles me, though, is a hyperbola as a decision boundary
since a hyperbola partitions the plane into three regions
(two of which are congruent). The joint density surfaces are
a (possibly) flattened bell and a circular bell, and
so one of these bells subsumes the other in two non-contiguous regions
of the plane: I just can't visualize it. Perhaps someone will
create a nice illustration....
Best Answer
It means that in the population, the prior probability that a sample belongs to either of the two classes is equal. Think of a fair coin; at each toss of the coin there is a 0.5 probability that the toss will result in a head or a tail.
If the class prior probabilities were not equal, that would indicate that in the population one of the two classes was more prevalent than the other.