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 $\alpha \in R$, does not meet the hyperbola ${x}^{2}-9{y}^{2}=9$, then a possible value of $\alpha $ is :

Accepted Answer

$\begin{gathered} \alpha x+2y=1,\alpha \in R \ {x}^{2}-9{y}^{2}=9 \ y=\frac{1-\alpha x}{2} \ 4{x}^{2}-9(1+{\alpha }^{2}{x}^{2}-2\alpha x)=36 \end{gathered}$

$\begin{gathered} \Delta0 \ {\alpha }^{2}>0.55 \ \alpha =0.8 \end{gathered}$

Solutions · 1