If the eccentric angles of two points P and Q on the ellipse `x^2/a^2+y^2/b^2` are `alpha,beta` such that `alpha +beta=pi/2`, then the locus of the point of intersection of the normals at P and Q is

Options

A. ax+by=0
B. ax-by=0
C. x+y=0
D. x+y=a+b

Answer

To solve the problem, we need to find the locus of the point of intersection of the normals at points P and Q on the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where the eccentric angles of points P and Q are \( \alpha \) and \( \beta \) respectively, with the condition that \( \alpha + \beta = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Identify the Coordinates of Points P and Q:**
The coordinates of point P corresponding to the eccentric angle \( \alpha \) are given by:
\[ P = (a \cos \alpha, b \sin \alpha) \]
The coordinates of point Q corresponding to the eccentric angle \( \beta \) are given by:
\[ Q = (a \cos \beta, b \sin \beta) \]

2. **Write the Equation of the Normal at Point P:**
The equation of the normal to the ellipse at point \( P(x_1, y_1) \) is:
\[ \frac{a^2 x}{a \cos \alpha} - \frac{b^2 y}{b \sin \alpha} = a^2 - b^2 \]
Simplifying this, we get:
\[ \frac{a^2 x}{\cos \alpha} - \frac{b^2 y}{\sin \alpha} = a^2 - b^2 \tag{1} \]

3. **Write the Equation of the Normal at Point Q:**
Similarly, for point Q, the equation of the normal is:
\[ \frac{a^2 x}{a \cos \beta} - \frac{b^2 y}{b \sin \beta} = a^2 - b^2 \]
Simplifying this, we obtain:
\[ \frac{a^2 x}{\cos \beta} - \frac{b^2 y}{\sin \beta} = a^2 - b^2 \tag{2} \]

4. **Substituting \( \beta \) with \( \frac{\pi}{2} - \alpha \):**
Since \( \beta = \frac{\pi}{2} - \alpha \), we can substitute this into equation (2):
\[ \frac{a^2 x}{\cos(\frac{\pi}{2} - \alpha)} - \frac{b^2 y}{\sin(\frac{\pi}{2} - \alpha)} = a^2 - b^2 \]
This simplifies to:
\[ \frac{a^2 x}{\sin \alpha} - \frac{b^2 y}{\cos \alpha} = a^2 - b^2 \tag{3} \]

5. **Finding the Intersection of the Normals:**
Now, we have two equations (1) and (3). To find the intersection point, we can solve these equations simultaneously.

From equation (1):
\[ a^2 x \sin \alpha - b^2 y \cos \alpha = (a^2 - b^2) \sin \alpha \tag{4} \]
From equation (3):
\[ a^2 x \sin \alpha - b^2 y \cos \alpha = (a^2 - b^2) \cos \alpha \tag{5} \]

6. **Equating and Simplifying:**
Setting equations (4) and (5) equal gives:
\[ (a^2 - b^2) \sin \alpha = (a^2 - b^2) \cos \alpha \]
If \( a^2 \neq b^2 \), we can divide both sides by \( a^2 - b^2 \):
\[ \sin \alpha = \cos \alpha \]
This implies:
\[ \tan \alpha = 1 \quad \Rightarrow \quad \alpha = \frac{\pi}{4} \]

7. **Finding the Locus:**
The locus of the intersection of the normals can be expressed as:
\[ ax + by = 0 \]
Hence, the locus of the point of intersection of the normals at points P and Q is:
\[ ax + by = 0 \]

### Final Answer:
The locus of the point of intersection of the normals at points P and Q is given by:
\[ ax + by = 0 \]