Explanation

(b)
We have, $x^{2} - {3y}^{2} = 0$…(i)
and $x = 4$ …(ii)
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From Eqs. (i) and (ii), we get
$y^{2} = \frac{16}{3}$
$\Rightarrow y = \pm \frac{4}{\sqrt{3}}$
$\therefore$ Three sides of triangle are $x - \sqrt{3}y = 0,\ x + \sqrt{3}y = 0$ and
$x - 4 = 0ie,\ OP = OQ = PQ = \frac{8}{\sqrt{3}}$
$\therefore$ Triangle is an equilateral triangle