If the curve `x^2 + 2y^2 = 2` intersects the line `x + y = 1` at two points P and Q, then the angle subtended by the line segment PQ at the origin is :

Options

A. `pi/2 + tan^(-1)(1/3)`
B. `pi/2 - tan^(-1)(1/3)`
C. `pi/2 - tan^(-1)(1/4)`
D. `pi/2 + tan^(-1)(1/4)`

Answer

To solve the problem of finding the angle subtended by the line segment PQ at the origin, where P and Q are the intersection points of the curve \(x^2 + 2y^2 = 2\) and the line \(x + y = 1\), we can follow these steps:

### Step 1: Substitute the line equation into the curve equation
We start with the line equation:
\[ x + y = 1 \implies y = 1 - x \]
Now, substitute \(y\) into the curve equation:
\[ x^2 + 2(1 - x)^2 = 2 \]

### Step 2: Expand and simplify the equation
Expanding the equation:
\[ x^2 + 2(1 - 2x + x^2) = 2 \]
\[ x^2 + 2 - 4x + 2x^2 = 2 \]
Combine like terms:
\[ 3x^2 - 4x + 2 - 2 = 0 \]
\[ 3x^2 - 4x = 0 \]

### Step 3: Factor the quadratic equation
Factoring out \(x\):
\[ x(3x - 4) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{or} \quad 3x - 4 = 0 \implies x = \frac{4}{3} \]

### Step 4: Find corresponding y-values
For \(x = 0\):
\[ y = 1 - 0 = 1 \quad \Rightarrow \quad P(0, 1) \]
For \(x = \frac{4}{3}\):
\[ y = 1 - \frac{4}{3} = -\frac{1}{3} \quad \Rightarrow \quad Q\left(\frac{4}{3}, -\frac{1}{3}\right) \]

### Step 5: Find the slopes of OP and OQ
The slope of OP (from the origin to point P):
\[ m_P = \frac{1 - 0}{0 - 0} = \text{undefined (vertical line)} \]
The slope of OQ (from the origin to point Q):
\[ m_Q = \frac{-\frac{1}{3} - 0}{\frac{4}{3} - 0} = -\frac{1}{4} \]

### Step 6: Calculate the angle between the two lines
Using the formula for the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\):
\[ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
Substituting \(m_1 = \text{undefined}\) (vertical line) and \(m_2 = -\frac{1}{4}\):
The angle between a vertical line and a line with slope \(m\) is given by:
\[ \theta = \tan^{-1}(|m|) = \tan^{-1}\left(\frac{1}{4}\right) \]

### Step 7: Conclusion
The angle subtended by the line segment PQ at the origin is:
\[ \theta = 90^\circ + \tan^{-1}\left(\frac{1}{4}\right) \]