A line through the point $(-a,0)$ cuts from the second quadrant a triangular region with area $T$. The equation of the line is:
$\textbf{(A)}\ 2Tx+a^2y+2aT=0 \qquad \textbf{(B)}\ 2Tx-a^2y+2aT=0 \qquad \textbf{(C)}\ 2Tx+a^2y-2aT=0 \qquad \\ \textbf{(D)}\ 2Tx-a^2y-2aT=0 \qquad \textbf{(E)}\ \text{none of these}$
The right triangle has area $T$ and base $a$, so the height $h$ satisfies $\frac{1}{2}ha = T$.  This means $h = \frac{2T}{a}$.  Becuase the triangle is in the second quadrant, the coordinates are the origin, $(-a, 0)$, and $(0, h)$, which means the third coordinate is $(0, \frac{2T}{a})$.
So, we want the equation of a line though $(-a, 0)$ and $(0, \frac{2T}{a})$.  The slope is $\frac{\frac{2T}{a} - 0}{0 - (-a)}$, which simplifies to $\frac{2T}{a^2}$.
The y-intercept is $(0, \frac{2T}{a})$, so the line in slope-intercept form is $y = mx + b$, or:

$y = \frac{2T}{a^2}x + \frac{2T}{a}$
All the solutions have a positive x-coefficient and no fractions, so we clear the fractions by multiplying by $a^2$ and move the $y$ term to the right to get:
$0 = 2Tx + 2aT - a^2y$
This is equivalent to option $\boxed{\textbf{(B)}}$