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

Question: 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 given equations of the ellipses are $x^2 + 4y^2 = 1$ and $4x^2 + y^2 = 1$.

To find the point of intersection, we equate the two expressions:

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

$3x^2 = 3y^2 \Rightarrow x = y$ (since $P$ lies in the first quadrant)

Substituting $x = y$ into the first ellipse equation:

$x^2 + 4x^2 = 1 \Rightarrow 5x^2 = 1 \Rightarrow x = \dfrac{1}{\sqrt{5}}$

Thus, the point of intersection is $P\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}\right)$.

Differentiating $x^2 + 4y^2 = 1$ with respect to $x$ gives the slope of the tangent to the first ellipse:

$2x + 8y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{4y}$

At $P\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}\right)$, the slope is $m_1 = -\dfrac{1}{4}$.

Differentiating $4x^2 + y^2 = 1$ with respect to $x$ gives the slope of the tangent to the second ellipse:

$8x + 2y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{4x}{y}$

At $P\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}\right)$, the slope is $m_2 = -4$.

The acute angle $\theta$ between the tangents is given by:

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

$\tan\theta = \left| \dfrac{-\dfrac{1}{4} - (-4)}{1 + \left(-\dfrac{1}{4}\right)(-4)} \right| = \left| \dfrac{\dfrac{15}{4}}{2} \right| = \dfrac{15}{8}$

Therefore, the value of $4\tan\theta$ is:

$4\tan\theta = 4 \times \dfrac{15}{8} = \dfrac{15}{2} = 7.5$

Answer: $7.5$