Three charges $-q_1,+q_2$ and  $-q_3$ are placed as shown in the figure. The x-component of the force on $-q_1$ is proportional to 

F₂ = Force applied by  $q_2$  on $-q_1$
F₃ = Force applied by $(-q_3)$  on – $-q_1$
x-component of Net force on  $-q_1$ is 

$F_x=F_2+F_3\sin \theta =k\frac{q_1{q}_2}{b^2}+k\cdot \frac{q_1{q}_3}{a^2}\sin \theta$
$\Rightarrow F_x=k⥄\left[\frac{q_1{q}_2}{b^2}+\frac{q_1{q}_3}{a^2}\sin \text{θ}\right]$

$F_x=k\cdot q_1⥄\left[\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \text{θ}\right]\Rightarrow$

$F_x\propto ⥄\left(\frac{q_2}{b^2}+\frac{q_3}{a^2}\sin \text{θ}\right)$