The locus of point of intersection of tangents to an ellipse `x^2/a^2+y^2/b^2=1` at two points the sum of whose eccentric angles is constant is

Options

A. parabola
B. circle
C. ellipse
D. straight line

Explanation

To find the locus of the point of intersection of tangents to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at two points where the sum of their eccentric angles is constant, we can follow these steps:

### Step 1: Define the Eccentric Angles
Let the eccentric angles of the two points on the ellipse be \( \theta_1 \) and \( \theta_2 \). We are given that the sum of these angles is constant, i.e., \( \theta_1 + \theta_2 = k \), where \( k \) is a constant.

### Step 2: Write the Equations of the Tangents
The equations of the tangents to the ellipse at points corresponding to the angles \( \theta_1 \) and \( \theta_2 \) are given by:
1. \( \frac{x}{a} \cos \theta_1 + \frac{y}{b} \sin \theta_1 = 1 \) (Equation 1)
2. \( \frac{x}{a} \cos \theta_2 + \frac{y}{b} \sin \theta_2 = 1 \) (Equation 2)

### Step 3: Solve for the Point of Intersection
To find the point of intersection of these two tangents, we can solve the two equations simultaneously.

From Equation 1:
\[ y = \frac{b}{\sin \theta_1} \left(1 - \frac{x \cos \theta_1}{a}\right) \]

Substituting this expression for \( y \) into Equation 2:
\[ \frac{x}{a} \cos \theta_2 + \frac{b}{\sin \theta_1} \left(1 - \frac{x \cos \theta_1}{a}\right) \sin \theta_2 = 1 \]

This will yield a relationship between \( x \) and \( \theta_1, \theta_2 \).

### Step 4: Express \( x \) and \( y \) in terms of \( k \)
Using the relation \( \theta_2 = k - \theta_1 \), we can express \( x \) and \( y \) in terms of \( k \) and \( \theta_1 \):
\[ x = a \cos \left(\frac{k}{2}\right) \sec \left(\frac{\theta_1 - (k - \theta_1)}{2}\right) \]
\[ y = b \sin \left(\frac{k}{2}\right) \sec \left(\frac{\theta_1 - (k - \theta_1)}{2}\right) \]

### Step 5: Find the Locus
The point of intersection \( (x_1, y_1) \) can be expressed as:
\[ x_1 = a \cos \left(\frac{k}{2}\right) \sec \left(\frac{\theta_1 - (k - \theta_1)}{2}\right) \]
\[ y_1 = b \sin \left(\frac{k}{2}\right) \sec \left(\frac{\theta_1 - (k - \theta_1)}{2}\right) \]

Now, we can derive the relationship between \( x_1 \) and \( y_1 \):
\[ \frac{y_1}{x_1} = \frac{b \sin \left(\frac{k}{2}\right)}{a \cos \left(\frac{k}{2}\right)} = \frac{b}{a} \tan \left(\frac{k}{2}\right) \]

### Step 6: Conclusion
The locus of the point of intersection of the tangents is a straight line given by:
\[ y = \frac{b}{a} \tan \left(\frac{k}{2}\right) x \]