An ellipse is inscribed in a reactangle and the angle between the diagonals of the reactangle is `tan^(-1)(2sqrt(2))`, then find the ecentricity of the ellipse

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Problem
We have an ellipse inscribed in a rectangle, and we are given the angle between the diagonals of the rectangle as \( \tan^{-1}(2\sqrt{2}) \). We need to find the eccentricity of the ellipse.

### Step 2: Set Up the Ellipse Equation
The standard form of the ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 3: Analyze the Rectangle and Its Diagonals
Let’s denote the angle between the diagonals of the rectangle as \( 2\theta \). From the problem, we have:

\[ \tan(2\theta) = 2\sqrt{2} \]

### Step 4: Use the Double Angle Formula
Using the double angle formula for tangent:

\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]

Setting this equal to \( 2\sqrt{2} \):

\[ \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = 2\sqrt{2} \]

### Step 5: Rearranging the Equation
Cross-multiplying gives:

\[ 2\tan(\theta) = 2\sqrt{2}(1 - \tan^2(\theta)) \]

This simplifies to:

\[ \tan(\theta) = \sqrt{2}(1 - \tan^2(\theta)) \]

### Step 6: Form a Quadratic Equation
Rearranging leads to:

\[ \sqrt{2}\tan^2(\theta) + \tan(\theta) - \sqrt{2} = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( \tan(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = \sqrt{2}, b = 1, c = -\sqrt{2} \):

\[ \tan(\theta) = \frac{-1 \pm \sqrt{1 + 8}}{2\sqrt{2}} = \frac{-1 \pm 3}{2\sqrt{2}} \]

This gives us two potential solutions:

1. \( \tan(\theta) = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \)
2. \( \tan(\theta) = \frac{-4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} \) (not valid since \( \tan(\theta) \) must be positive)

Thus, we have:

\[ \tan(\theta) = \frac{1}{\sqrt{2}} \]

### Step 8: Relate \( a \) and \( b \)
Since \( \tan(\theta) = \frac{b}{a} \):

\[ \frac{b}{a} = \frac{1}{\sqrt{2}} \implies b = \frac{a}{\sqrt{2}} \implies a = b\sqrt{2} \]

### Step 9: Calculate the Eccentricity
The eccentricity \( e \) of the ellipse is given by:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting \( b = \frac{a}{\sqrt{2}} \):

\[ e = \sqrt{1 - \frac{\left(\frac{a}{\sqrt{2}}\right)^2}{a^2}} = \sqrt{1 - \frac{a^2/2}{a^2}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
The eccentricity of the ellipse is:

\[ \boxed{\frac{1}{\sqrt{2}}} \]