Consider the ellipses given by $$x^2+4y^2=1$$ and $$4x^2+y^2=1$$.

Let $$P$$ be the point in the first quadrant where the given ellipses intersect. If $$\theta$$ is the acute angle between the tangents to the given ellipses at the point $$P$$, then the value of $$4\tan\theta$$ is ___.

The point of intersection of the two ellipses satisfies

$$x^2+4y^2=1$$ and $$4x^2+y^2=1$$

Subtracting,

$$3x^2-3y^2=0$$

$$x^2=y^2$$

Since $$P$$ lies in the first quadrant,

$$x=y$$

Substituting into

$$x^2+4y^2=1$$ gives $$5x^2=1$$

$$x=y=\frac1{\sqrt5}$$

Hence,

$$P=\left(\frac1{\sqrt5},\frac1{\sqrt5}\right)$$

For the ellipse

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

differentiating implicitly,

$$2x+8y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x}{4y}$$

At $$P$$,

$$m_1=-\frac14$$

For the ellipse

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

differentiating implicitly,

$$8x+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{4x}{y}$$

At $$P$$,

$$m_2=-4$$

The angle between the tangents is given by

$$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$$

Substituting,

$$\tan\theta=\left|\frac{-4+\frac14}{1+1}\right|$$

$$=\frac{\frac{15}{4}}{2}$$

$$=\frac{15}{8}$$

Therefore,

$$4\tan\theta=4\cdot\frac{15}{8}$$

$$=\frac{15}{2}$$

Hence,

$$\boxed{\frac{15}{2}}$$