Consider the ellipses given by x² + 4y² = 1 and 4x² + y² = 1.
Let P be the point in the first quadrant where the given ellipses intersect. If θ is the acute angle between the tangents to the given ellipses at the point P, then the value of 4 tanθ is .

The correct answer is 7.5.

Step 1: Understand the given equations of ellipses
We are given two ellipses:

First ellipse: $x^2+4y^2=1$   --- (Equation 1)

Second ellipse: $4x^2+y^2=1$   --- (Equation 2)

Step 2: Find the point of intersection P in the first quadrant
Adding Equation 1 and Equation 2 gives:

$5x^2+5y^2=2$

$x^2+y^2=\frac{2}{5}$

Subtracting Equation 2 from 4 times Equation 1 gives:
$4(x^2+4y^2)-(4x^2+y^2)=4\left(1\right)-1$
$15y^2=3$ ⇒ $y^2=\frac{1}{5}$

Substituting $y^2=\frac{1}{5}$ back into $x^2+4y^2=1$:
$x^2+4\left(\frac{1}{5}\right)=1$ ⇒ $x^2=\frac{1}{5}$

Since point P is in the first quadrant, both x and y coordinates must be positive:
$P=(\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}})$

Step 3: Find the slopes of the tangents at point P
Differentiating Equation 1 with respect to x:
$2x+8y\frac{dy}{dx}=0$ ⇒ $m_1=\frac{dy}{dx}=-\frac{x}{4y}$

At point P$(\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}})$:
$m_1=-\frac{1}{4}$

Differentiating Equation 2 with respect to x:
$8x+2y\frac{dy}{dx}=0$ ⇒ $m_2=\frac{dy}{dx}=-\frac{4x}{y}$

At point P$(\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}})$:
$m_2=-4$

Step 4: Calculate tan θ and 4 tan θ
The acute angle θ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Substituting $m_1=-\frac{1}{4}$ and $m_2=-4$:

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

Now, calculating the value of 4 tan θ:

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

Thus, the value of 4 tan θ is 7.5.