If the line x+2y=1, where R, does not meet the hyperbola x^2-9y^2=9, then a possible value…

Question

If the line $\alpha x+2y=1$, where $\alpha\in\mathbb R$, does not meet the hyperbola $x^2-9y^2=9$, then a possible value of $\alpha$ is:

Accepted Answer

Substituting $y=\dfrac{1-\alpha x}{2}$ into $x^2-9y^2=9$: $$4x^2-9(1-\alpha x)^2=36 \;\Rightarrow\; (4-9\alpha^2)x^2+18\alpha x-45=0.$$ For the line to miss the hyperbola, this quadratic must have no real roots, i.e. discriminant $<0$: $$(18\alpha)^2+180(4-9\alpha^2)<0 \;\Rightarrow\; -1296\alpha^2+720<0 \;\Rightarrow\; \alpha^2>\frac{5}{9}\approx0.556.$$ Among the options only $\alpha=0.8$ satisfies $\alpha^2>0.556$.

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