If `tantheta_1.tantheta_2=a^2/b^2` then the chord Joining two points `theta_1 and theta_2` on the ellipse `x^2/a^2+y^2/b^2=1` will subtend a right angle at (A) focus (B) centre (C) end of the major axis (D) end of the major axis

Options

A. focus
B. center
C. end of major axis
D. end of minor axis

Answer

To solve the problem, we need to determine at which point the chord joining two points on the ellipse subtends a right angle, given that \( \tan \theta_1 \cdot \tan \theta_2 = \frac{a^2}{b^2} \).

### Step-by-Step Solution:

1. **Understanding the Ellipse**:
The equation of the ellipse is given as:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The points on the ellipse corresponding to angles \( \theta_1 \) and \( \theta_2 \) can be represented as:
\[ P = (a \cos \theta_1, b \sin \theta_1) \quad \text{and} \quad Q = (a \cos \theta_2, b \sin \theta_2) \]

2. **Finding the Slopes**:
The slope of the line segment \( OP \) (from the origin to point \( P \)) is given by:
\[ \text{slope of } OP = \frac{b \sin \theta_1 - 0}{a \cos \theta_1 - 0} = \frac{b}{a} \tan \theta_1 \]
Similarly, the slope of the line segment \( OQ \) (from the origin to point \( Q \)) is:
\[ \text{slope of } OQ = \frac{b \sin \theta_2 - 0}{a \cos \theta_2 - 0} = \frac{b}{a} \tan \theta_2 \]

3. **Using the Given Condition**:
We are given that:
\[ \tan \theta_1 \cdot \tan \theta_2 = \frac{a^2}{b^2} \]

4. **Multiplying the Slopes**:
The product of the slopes of \( OP \) and \( OQ \) is:
\[ \left( \frac{b}{a} \tan \theta_1 \right) \cdot \left( \frac{b}{a} \tan \theta_2 \right) = \frac{b^2}{a^2} \tan \theta_1 \tan \theta_2 \]
Substituting the given condition:
\[ = \frac{b^2}{a^2} \cdot \frac{a^2}{b^2} = 1 \]

5. **Conclusion about Perpendicularity**:
Since the product of the slopes is equal to \( -1 \) (as we see that the slopes are negative reciprocals), this indicates that the lines \( OP \) and \( OQ \) are perpendicular. Therefore, the chord \( PQ \) subtends a right angle at the origin (the center of the ellipse).

### Final Answer:
The chord joining the two points \( \theta_1 \) and \( \theta_2 \) on the ellipse subtends a right angle at the **center** of the ellipse.

**Correct Option**: (B) center