NIMCET 2014 Question Paper with solutions

Let the vertex P be (α, β), so that 0 is (α, -β) such that
ΔOPQ is equilateral

OP = OQ = PQ = 2β
$\sqrt{{\mathit{\alpha }}^2+{\mathit{\beta }}^2}$ = 2β
or ${\mathit{\alpha }}^2$ = $3{\mathit{\beta }}^2$ ... (i)
Hence, point (a, P) lies on hyperbola
∴ $\frac{{\mathit{\alpha }}^2}{{\mathit{a}}^2}-\frac{{\mathit{\beta }}^2}{{\mathit{b}}^2}$ = 1
⇒ $\frac{3{\mathit{\beta }}^2}{{\mathit{a}}^2}-\frac{{\mathit{\beta }}^2}{{\mathit{b}}^2}$ = 1
⇒ $\frac{3{\mathit{\beta }}^2}{{\mathit{a}}^2}$ - 1 = $\frac{{\mathit{\beta }}^2}{{\mathit{b}}^2}$ = ± 1 + ve > 0
⇒ $\frac{{\mathit{b}}^2}{{\mathit{a}}^2}$ > $\frac{1}{3}$ ⇒ ${\mathit{b}}^2$ > $\frac{{\mathit{a}}^2}{3}$