If the electric field E is plotted, with distance from $q_{1}$ along the line joining two charges $q_{1}$ and $q_{2}$ , then (E is positive along $+ ve x - axis)$ it looks as shown in figure. From the plot, we can say that

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$q_{1} and q_{2}$ both are negative

$|\frac{q_{1}}{q_{2}}| = \frac{1}{4}$

$|\frac{q_{1}}{q_{2}}| = 4$

$q_{1}$ is positive and $q_{2}$ is negative

answer is A, B.

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Detailed Solution

To the immediate right side neighbourhood of $q_{1}$ , electric field (force on a unit positive charge) is highly negative. It means $q_{1}$ is negative.
To the immediate left side neighbourhood of $q_{2}$ , electric field (force on a unit positive charge) is highly positive. It means $q_{2}$ is negative.
$q_{1}$ and $q_{2}$ both are negative Option (a) is correct, option (c) is wrong
At the null point,

$\frac{k q_{1}}{{(\frac{2 a}{3})}^{2}} = \frac{k q_{2}}{{(\frac{a}{3})}^{2}}$ $\Rightarrow \frac{q_{1}}{q_{2}} = 4$ Option (b) is correct, option (d) is wrong