Let a line $$L_1$$ be tangent to the hyperbola $$\frac{x^2}{16} - \frac{y^2}{4} = 1$$ and let $$L_2$$ be the line passing through the origin and perpendicular to $$L_1$$. If the locus of the point of intersection of $$L_1$$ and $$L_2$$ is $$(x^2 + y^2)^2 = \alpha x^2 + \beta y^2$$, then $$\alpha + \beta$$ is equal to ______

Let $$L_1$$ be a tangent to the hyperbola $$\frac{x^2}{16} - \frac{y^2}{4} = 1$$, and $$L_2$$ be the line through the origin perpendicular to $$L_1$$. We need to find the locus of intersection of $$L_1$$ and $$L_2$$ in the form $$(x^2 + y^2)^2 = \alpha x^2 + \beta y^2$$, and compute $$\alpha + \beta$$.

First, for the hyperbola $$\frac{x^2}{16} - \frac{y^2}{4} = 1$$ (where $$a^2 = 16, b^2 = 4$$), a tangent with slope $$m$$ is given by the line $$y = mx + c$$, subject to the condition $$c^2 = 16m^2 - 4$$.

Next, the line through the origin perpendicular to $$L_1$$ has slope $$-\tfrac{1}{m}$$, so it is described by

$$L_2: y = -\frac{x}{m}$$.

Now let $$(h,k)$$ be the point of intersection of $$L_1$$ and $$L_2$$. From $$L_2$$ we have $$k = -\tfrac{h}{m}$$, or equivalently $$h = -mk$$. Substituting into $$L_1$$ gives

$$k = m(-mk) + c \implies k(1 + m^2) = c \implies k = \frac{c}{1 + m^2},$$

and hence

$$h = -mk = \frac{-mc}{1 + m^2}.$$

Since

$$h^2 + k^2 = \frac{m^2c^2 + c^2}{(1+m^2)^2} = \frac{c^2}{1+m^2},$$

we can express $$m^2$$ and $$c^2$$ in terms of $$h, k$$. From $$h = -mk$$ we obtain $$m = -\frac{h}{k}$$, so $$m^2 = \frac{h^2}{k^2}$$. Substituting into the previous relation yields

$$c^2 = (h^2 + k^2)(1 + m^2) = (h^2 + k^2)\cdot\frac{h^2 + k^2}{k^2} = \frac{(h^2 + k^2)^2}{k^2}.$$

Substituting these into the tangency condition $$c^2 = 16m^2 - 4$$ gives

$$\frac{(h^2 + k^2)^2}{k^2} = 16\cdot\frac{h^2}{k^2} - 4.$$

Multiplying both sides by $$k^2$$ leads to

$$ (h^2 + k^2)^2 = 16h^2 - 4k^2. $$

Comparing with the desired form $$(x^2 + y^2)^2 = \alpha x^2 + \beta y^2$$ shows that $$\alpha = 16$$ and $$\beta = -4$$. Therefore,

$$\alpha + \beta = 16 + (-4) = 12.$$

Hence the answer is $$12$$.