If normal at any point P to ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1(a > b)` meet the x & y axes at A and B respectively. Such that `(PA)/(PB) = 3/4` , then eccentricity of the ellipse is:

Options

A. `1/3`
B. `1/2`
C. `1/4`
D. `1/9`

Explanation

To solve the problem, we need to find the eccentricity of the ellipse given the condition that the ratio of the distances from point P to points A and B is \( \frac{3}{4} \).

### Step-by-Step Solution:

1. **Understanding the ellipse**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a > b \). The major axis is along the x-axis, and the minor axis is along the y-axis.

2. **Finding the coordinates of point P**:
Let point P on the ellipse be represented in parametric form as:
\[ P(a \cos \theta, b \sin \theta) \]

3. **Equation of the normal at point P**:
The equation of the normal to the ellipse at point P can be derived using the formula:
\[ y - b \sin \theta = -\frac{b^2 \cos \theta}{a^2 \sin \theta} (x - a \cos \theta) \]
Rearranging gives:
\[ \frac{b^2 \cos \theta}{a^2 \sin \theta} x + y = b \sin \theta + \frac{b^2 \cos \theta}{a^2 \sin \theta} a \cos \theta \]

4. **Finding intersection with the x-axis (A)**:
Set \( y = 0 \) to find the x-intercept (point A):
\[ 0 = b \sin \theta + \frac{b^2 \cos \theta}{a^2 \sin \theta} a \cos \theta - \frac{b^2 \cos \theta}{a^2 \sin \theta} x \]
Solving for \( x \):
\[ x_A = \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{b \sin \theta} \]

5. **Finding intersection with the y-axis (B)**:
Set \( x = 0 \) to find the y-intercept (point B):
\[ y_B = b \sin \theta - \frac{b^2 \cos \theta}{a^2 \sin \theta} a \cos \theta \]

6. **Using the given ratio**:
We know that:
\[ \frac{PA}{PB} = \frac{3}{4} \]
This gives us:
\[ PA = \sqrt{(x_A - a \cos \theta)^2 + (0 - b \sin \theta)^2} \]
\[ PB = \sqrt{(0 - a \cos \theta)^2 + (y_B - b \sin \theta)^2} \]

7. **Setting up the equation**:
Substitute \( PA \) and \( PB \) into the ratio:
\[ \frac{PA}{PB} = \frac{3}{4} \]
Squaring both sides and simplifying will yield a relationship between \( a \) and \( b \).

8. **Finding eccentricity**:
The eccentricity \( e \) of the ellipse is defined as:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Using the relationship derived from the ratio, we can solve for \( e \).

9. **Final calculation**:
After simplification, we find that:
\[ 3a^2 = 4b^2 \quad \Rightarrow \quad \frac{b^2}{a^2} = \frac{3}{4} \]
Therefore, substituting into the eccentricity formula:
\[ e = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Conclusion:
The eccentricity of the ellipse is:
\[ \boxed{\frac{1}{2}} \]