If from (1, $\alpha$) two tangents are drawn on exactly one branch of the hyperbola $\frac{x^2}{4} -\frac{y^2}{1} = 1$ the alpha belongs to

Question

If from (1, $\alpha$) two tangents are drawn on exactly one branch of the hyperbola $$\frac{x^2}{4} -\frac{y^2}{1} = 1$$ the alpha belongs to

As far as I can see two tangents can be drawn to only one branch if the point lies inside the branch opposite to it (the white area which is technically the outside but it looks inside ).

(1, alpha) lies in the blue region so we should be able to draw 2 tangents to both of the branches.

If it helps the range of alpha is given as $( -1/2, 1/2)$

Answer 1

The tangent to any hyperbola is of the form $y = mx ± \sqrt{a^2m^2 - b^2}$. Here, it will be of the form, $y = mx ± \sqrt{4m^2 - 1}$. Squaring both sides we get a quadratic equation,

$m^2(x^2 - 4) - (2xy)m + y^2 + 1 = 0$

The equation must have exactly two roots, $D>0$. Substitute $x=1$ and find the inequality for $y$. This works because from that point you can draw tangents to only one branch.

Answer 2

Hint: With $$y=mx+n$$ and $$P(1,\alpha)$$ we get $$\frac{x^2}{4}-(m(x-1)+\alpha)^2=1$$ Solve this equation for $x$