$v = - mx + {v_0}$…..(i)       [where $m = \tan \theta = \frac{{{v_0}}}{{{x_0}}}$]

By differentiating with respect to time we get $\frac{{dv}}{{dt}} = - m\frac{{dx}}{{dt}} = - mv$

Now substituting the value of $v$ from eq. (i) we get $\frac{{dv}}{{dt}} = - m[ - mx + {v_0}] = {m^2}x - m{v_0}$

$\therefore a = {m^2}x - m{v_0}$

i.e. the graph between $a$ and $x$ should have positive slope but negative intercept on $a-$axis. So graph $(a) $ is correct.