Question

Let the curve \(x^2 + 2y^2 = 2\) intersect the line \(x + y = 1\) at two points \(P\) and \(Q\) and \(O\) be the origin. If \(\theta\) is the acute angle between the lines \(OP\) and \(OQ\), then \(\tan \theta =\)

• A. \(\frac{1}{4}\)
• B. \(4\)
• C. \(\sqrt{3}\)
• D. \(\frac{1}{\sqrt{3}}\)

Hint:

To find angle between vectors, use the identity: \(\tan \theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}}\) for vectors from origin.

Answer 1

The Correct Option is B

We are given: \[ x + y = 1 \quad \text{(1)}
x^2 + 2y^2 = 2 \quad \text{(2)} \] From (1), substitute \(x = 1 - y\) into (2): \[ (1 - y)^2 + 2y^2 = 2 \Rightarrow 1 - 2y + y^2 + 2y^2 = 2 \Rightarrow 3y^2 - 2y -1 = 0 \] Solving: \[ y = \frac{2 \pm \sqrt{(-2)^2 + 4(3)(1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \Rightarrow y = 1,\ -\frac{1}{3} \] So corresponding points are: \[ P = (0,1), \quad Q = \left(\frac{4}{3}, -\frac{1}{3} \right) \] Now find angle between \( \vec{OP} \) and \( \vec{OQ} \) using: \[ \tan \theta = \frac{|\vec{OP} \times \vec{OQ}|}{\vec{OP} \cdot \vec{OQ}}
\vec{OP} = (0,1),\ \vec{OQ} = \left( \frac{4}{3}, -\frac{1}{3} \right) \] Cross product magnitude: \[ |\vec{OP} \times \vec{OQ}| = |0 \cdot (-1/3) - 1 \cdot (4/3)| = \frac{4}{3} \] Dot product: \[ 0 \cdot (4/3) + 1 \cdot (-1/3) = -1/3 \] \[ \tan \theta = \left| \frac{4/3}{-1/3} \right| = 4 \]