Let a line $$L : 2x + y = k$$, $$k > 0$$ be a tangent to the hyperbola $$x^2 - y^2 = 3$$. If $$L$$ is also a tangent to the parabola $$y^2 = \alpha x$$, then $$\alpha$$ is equal to:

Write the hyperbola as $$\frac{x^2}{3}-\frac{y^2}{3}=1$$

So,

$$a^2=3,\qquad b^2=3$$

The line $$2x+y=k$$ becomes $$y=-2x+k$$

Therefore,

$$m=-2,\qquad c=k$$

For the line $$y=mx+c$$ to touch the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ the condition is

$$c^2=a^2m^2-b^2$$

Substituting the values,

$$k^2=3(-2)^2-3$$

$$=12-3=9$$

Since

$$k>0,$$ 

$$k=3$$

So the tangent line is $$y=-2x+3$$

Now compare the parabola $$y^2=\alpha x$$ with $$y^2=4ax$$

This gives $$4a=\alpha$$

For the parabola $$y^2=4ax,$$ the tangent with slope $$m$$ is $$y=mx+\frac{a}{m}$$

Comparing with $$y=-2x+3,$$

$$3=\frac{a}{-2}$$

$$a=-6$$

Therefore,

$$\alpha=4a=-24$$