If the length of the major axis intercepted between the tangent and normal at a point `P (a cos theta, b sin theta)` on the ellipse `(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1` is equal to the length of semi-major axis, then eccentricity of the ellipse is

Options

A. `(cos theta)/(sqrt(1-cos theta))`
B. `(sqrt(1-cos theta))/(cos theta)`
C. `(sqrt(1-cos theta))/(sin theta)`
D. `(sin theta)/(sqrt(1-sin theta))`

Explanation

To solve the problem, we need to find the eccentricity of the ellipse given that the length of the major axis intercepted between the tangent and normal at a point \( P(a \cos \theta, b \sin \theta) \) is equal to the length of the semi-major axis.

### Step-by-Step Solution:

1. **Identify the Ellipse Equation**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

2. **Find the Tangent at Point \( P \)**:
The point \( P \) on the ellipse is \( (a \cos \theta, b \sin \theta) \). The equation of the tangent at point \( P \) is given by:
\[ \frac{x \cdot \cos \theta}{a} + \frac{y \cdot \sin \theta}{b} = 1 \]

3. **Find the Intersection of Tangent with the X-axis**:
To find where the tangent intersects the x-axis, set \( y = 0 \) in the tangent equation:
\[ \frac{x \cdot \cos \theta}{a} = 1 \implies x = \frac{a}{\cos \theta} \]

4. **Find the Normal at Point \( P \)**:
The slope of the tangent at point \( P \) can be derived from the derivative of the ellipse equation. The normal at point \( P \) will have a slope that is the negative reciprocal of the tangent slope. The equation of the normal is:
\[ y - b \sin \theta = -\frac{b \cos \theta}{a \sin \theta}(x - a \cos \theta) \]

5. **Find the Intersection of Normal with the X-axis**:
Set \( y = 0 \) in the normal equation and solve for \( x \):
\[ 0 - b \sin \theta = -\frac{b \cos \theta}{a \sin \theta}(x - a \cos \theta) \]
Rearranging gives:
\[ b \sin \theta = \frac{b \cos \theta}{a \sin \theta}(x - a \cos \theta) \]
Solving for \( x \) leads to:
\[ x = \frac{a^2 - b^2}{a \cos \theta} + a \cos \theta \]

6. **Length of Major Axis Intercepted**:
The length of the major axis intercepted between the tangent and normal is given by:
\[ L = \left(\frac{a}{\cos \theta} - \left(\frac{a^2 - b^2}{a \cos \theta} + a \cos \theta\right)\right) \]
Simplifying this expression gives us the length \( L \).

7. **Set the Length Equal to Semi-Major Axis**:
According to the problem, this length \( L \) is equal to the semi-major axis \( a \):
\[ L = a \]

8. **Solve for Eccentricity**:
The eccentricity \( e \) of the ellipse is defined as:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
From the previous steps, we can derive a relationship involving \( \cos \theta \) and \( e \):
\[ e^2 \cos^2 \theta = 1 - \cos^2 \theta \]
Rearranging gives:
\[ e = \sqrt{\frac{1 - \cos^2 \theta}{\cos^2 \theta}} = \sqrt{\frac{\sin^2 \theta}{\cos^2 \theta}} = \tan \theta \]

### Final Answer:
The eccentricity of the ellipse is:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]