If two tangents drawn from a point $$(\alpha, \beta)$$ lying on the ellipse $$25x^2 + 4y^2 = 1$$ to the parabola $$y^2 = 4x$$ are such that the slope of one tangent is four times the other, then the value of $$(10\alpha + 5)^2 + (16\beta^2 + 50)^2$$ equals ______

Since the tangents are drawn from $$(\alpha,\beta)$$ to the parabola

$$y^2=4x,$$

the tangent with slope $$m$$ is

$$y=mx+\frac1m$$

As $$(\alpha,\beta)$$ lies on it,

$$\beta=\alpha m+\frac1m$$

$$\alpha m^2-\beta m+1=0$$

Let the slopes be $$m_1,m_2.$$

Then,

$$m_1+m_2=\frac{\beta}{\alpha},\qquad m_1m_2=\frac1\alpha$$

Given one slope is four times the other.

Let

$$m_1=4m_2$$

Then,

$$m_1+m_2=5m_2=\frac{\beta}{\alpha}$$

$$m_1m_2=4m_2^2=\frac1\alpha$$

From the first equation,

$$m_2=\frac{\beta}{5\alpha}$$

Substituting in the second,

$$4\left(\frac{\beta}{5\alpha}\right)^2=\frac1\alpha$$

$$\frac{4\beta^2}{25\alpha^2}=\frac1\alpha$$

$$4\beta^2=25\alpha$$

Since $$(\alpha,\beta)$$ lies on

$$25x^2+4y^2=1,$$

we get

$$25\alpha^2+4\beta^2=1$$

Using

$$4\beta^2=25\alpha,$$

$$25\alpha^2+25\alpha=1$$

$$25\alpha^2+25\alpha-1=0$$

Multiplying by $$4,$$

$$100\alpha^2+100\alpha-4=0$$

Adding $$25$$ on both sides,

$$100\alpha^2+100\alpha+25=29$$

$$=(10\alpha+5)^2$$

Hence,

$$(10\alpha+5)^2=29$$

Also,

$$16\beta^2=4(4\beta^2)=4(25\alpha)=100\alpha$$

Therefore,

$$16\beta^2+50=100\alpha+50=10(10\alpha+5)$$

Squaring,

$$(16\beta^2+50)^2=100(10\alpha+5)^2$$

$$=100(29)=2900$$

Thus,

$$(10\alpha+5)^2+(16\beta^2+50)^2$$

$$=29+2900$$

$$=2929$$

Hence, the required value is

$$\boxed{2929}$$.