If the maximum distance of normal to the ellipse `(x^(2))/(4)+(y^(2))/(b^(2))=1,b<2,` from the origin is `1`,then the eccentricity of the ellipse is :

Options

A. `(sqrt(3))/(4)`
B. `(sqrt(3))/(2)`
C. `(1)/(sqrt(2))`
D. `(1)/(2)`

Answer

To find the eccentricity of the ellipse given that the maximum distance of the normal to the ellipse from the origin is 1, we can follow these steps:

### Step 1: Write down the equation of the ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{4} + \frac{y^2}{b^2} = 1 \]
Here, \( a^2 = 4 \) and \( b^2 = b^2 \).

### Step 2: Write the equation of the normal to the ellipse
The equation of the normal to the ellipse at a point \( (x_0, y_0) \) can be expressed as:
\[ 2x_0 (x - x_0) - b^2 (y - y_0) = 4 - b^2 \]
This can be rearranged to:
\[ 2x_0 x - b^2 y = 4 - b^2 + 2x_0^2 \]

### Step 3: Calculate the distance from the origin to the normal
The distance \( D \) from the origin \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ D = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For our normal equation, we can identify \( A = 2x_0 \), \( B = -b^2 \), and \( C = 4 - b^2 + 2x_0^2 \). Thus, the distance from the origin is:
\[ D = \frac{|4 - b^2 + 2x_0^2|}{\sqrt{(2x_0)^2 + (-b^2)^2}} = \frac{|4 - b^2 + 2x_0^2|}{\sqrt{4x_0^2 + b^4}} \]

### Step 4: Find the maximum distance condition
We know that the maximum distance \( D \) is given as 1. Therefore, we set up the equation:
\[ \frac{|4 - b^2 + 2x_0^2|}{\sqrt{4x_0^2 + b^4}} = 1 \]
This implies:
\[ |4 - b^2 + 2x_0^2| = \sqrt{4x_0^2 + b^4} \]

### Step 5: Analyze the condition for maximum distance
To maximize \( D \), we need to minimize the denominator \( \sqrt{4x_0^2 + b^4} \). The minimum occurs when \( x_0 \) is chosen such that \( \tan^2 \theta = \frac{b}{2} \).

### Step 6: Substitute and solve for \( b \)
Substituting \( x_0 = 2 \) (since \( a = 2 \) for our ellipse):
\[ 4 - b^2 = b + 2 \]
Solving this gives:
\[ 4 - b^2 = b + 2 \implies b^2 + b - 2 = 0 \]
Factoring:
\[ (b - 1)(b + 2) = 0 \]
Thus, \( b = 1 \) (since \( b < 2 \)).

### Step 7: Find the eccentricity
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1^2}{2^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Final Answer
The eccentricity of the ellipse is:
\[ \frac{\sqrt{3}}{2} \]