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I have seen in the following reaction of azo coupling where coupling takes place at position 4 of $\alpha$-naphthol.

Coupling reaction of alpha-napthol

Now the electron density accumulation is similar at both position $4$ as well as $5$.

Mechanism of reaction

However, there are more resonance structures that can be drawn for position 5. It should also be sterically easier to attack position 5.

The only reason I can think of for position 4 over 5 is that $\ce{=OH+}$ has a -I effect, which may somewhat stabilize the carbanion. But at the same time due to large distance this may be very weak and may not efficiently stabilize the negative charge. Then why is it in the para and not the peri position that coupling preferentially occurs?

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The product at the para-position is more stable as it allows the compound to exist in the quinoid form, rather than only in the benzoid form:

tautomerism, their percentages, and ratio

This is important as in your case, if the substituent $\ce{X}$ is $\ce{H}$, then depending on the solvent, as much as 89% of it may exist in the quinoid form ($\bf{2}$).

References:

  1. Satoshi Kishimoto, Shinya Kitahara, Osamu Manabe, Hachiro Hiyama, “Studies on coupling reactions of 1-naphthol. 6. Tautomerism and dissociation of 4-arylazo-1-naphthols in various solvents,” J. Org. Chem. 1978, 43(20), 3882–3886 (https://doi.org/10.1021/jo00414a020).
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