`P_1(theta_1)` and `P_2(theta_2)` are two points on the ellipse `x^2/a^2+y^2/b^2=1` such that tan `theta_1` tan `theta_2 = (-a^2)/b^2`. The chord joining `P_1` and `P_2` of the ellipse subtends a right angle at the

Options

A. focus (ae,0)
B. focus (-ae, 0)
C. centre (0,0)
D. vertex (a,0)

Answer

To solve the problem, we need to find the points \( P_1(\theta_1) \) and \( P_2(\theta_2) \) on the ellipse defined by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We know that the product of the tangents of the angles corresponding to these points is given by:

\[ \tan \theta_1 \tan \theta_2 = -\frac{a^2}{b^2} \]

We will also show that the chord joining these two points subtends a right angle at the origin (the center of the ellipse).

### Step-by-Step Solution

1. **Parametric Representation of Points**:
The points \( P_1 \) and \( P_2 \) on the ellipse can be expressed in parametric form as:
\[ P_1 = (a \cos \theta_1, b \sin \theta_1) \]
\[ P_2 = (a \cos \theta_2, b \sin \theta_2) \]

2. **Finding Slopes**:
The slope of the line segment joining the origin \( O(0, 0) \) to point \( P_1 \) is:
\[ m_1 = \frac{b \sin \theta_1 - 0}{a \cos \theta_1 - 0} = \frac{b \sin \theta_1}{a \cos \theta_1} \]
Similarly, the slope of the line segment joining the origin to point \( P_2 \) is:
\[ m_2 = \frac{b \sin \theta_2}{a \cos \theta_2} \]

3. **Condition for Right Angle**:
For the chord \( P_1P_2 \) to subtend a right angle at the origin, the product of the slopes \( m_1 \) and \( m_2 \) must be equal to -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \left( \frac{b \sin \theta_1}{a \cos \theta_1} \right) \left( \frac{b \sin \theta_2}{a \cos \theta_2} \right) = -1 \]
Simplifying this gives:
\[ \frac{b^2 \sin \theta_1 \sin \theta_2}{a^2 \cos \theta_1 \cos \theta_2} = -1 \]
Rearranging leads to:
\[ b^2 \sin \theta_1 \sin \theta_2 = -a^2 \cos \theta_1 \cos \theta_2 \]

4. **Using the Given Condition**:
From the problem, we have:
\[ \tan \theta_1 \tan \theta_2 = -\frac{a^2}{b^2} \]
This implies:
\[ \frac{\sin \theta_1}{\cos \theta_1} \cdot \frac{\sin \theta_2}{\cos \theta_2} = -\frac{a^2}{b^2} \]
Rearranging gives:
\[ b^2 \sin \theta_1 \sin \theta_2 = -a^2 \cos \theta_1 \cos \theta_2 \]
This matches our earlier result, confirming that the condition for the right angle is satisfied.

5. **Conclusion**:
Since the slopes' product is -1, the chord \( P_1P_2 \) indeed subtends a right angle at the origin, which is the center of the ellipse.

### Final Answer:
The chord joining \( P_1 \) and \( P_2 \) subtends a right angle at the origin (the center of the ellipse).

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