Locate in Cartesian plane the geometric place of points $(x, y)$ for which equation in $ \alpha $ has two roots of same sign

Question

The equation

$$ \alpha ^ 2 - (x-2y) \alpha-xy + 1 = 0 $$

admits two roots of same sign. Locate in the Cartesian plane the geometric place of points $(x; y)$.

What I tried:

To have two roots you need $\Delta> 0$,

$ \Delta = (x-2y) ^ 2 - 4 (xy +1) = X ^ 2 + y + 4y ^ 2 + 4xy-4 = x ^ 2 + 4y ^ 2-4> 0 $

Ae of $x ^ 2 + 4y ^ 2-4 = 0,$ we have

$ \frac {x ^ 2} {4} + y ^ 2 = 1 $ (a graphical representation)

Answer 1

You also need to confine the region with the condition that the two roots have the same sign, which is $\alpha_1\alpha_2 = -xy+1 > 0$. Along with the condition of two real roots, establish the inequalities below,

$$ \frac {x ^ 2} {4} + y ^ 2 > 1 $$ $$ xy < 1$$

which jointly define the location of $(x,y)$, as shown by the shaded area in the graph below.