Question

Let \( \frac{x^2}{2} + \frac{y^2}{1} = 1 \) and \( y = x + 1 \) intersect each other at points A & B, then \( \angle AOB \) (where O is the centre of the ellipse) is:

• A. \( \frac{\pi}{2} + \tan^{-1} \frac{1}{4} \)
• B. \( \frac{\pi}{2} + \tan^{-1} \frac{1}{3} \)
• C. \( \frac{\pi}{2} + \tan^{-1} \frac{1}{2} \)
• D. \( \frac{\pi}{4} + \tan^{-1} \frac{1}{2} \)

Hint:

When finding angles in ellipses and lines, use the properties of tangents and the geometry of the ellipse to simplify the calculations.

Answer 1

The Correct Option is A

To find the angle \( \angle AOB \) where points \( A \) and \( B \) are the points of intersection of the ellipse and the line, follow these steps:

1. First, consider the equations of the ellipse and the line. The equation of the ellipse is \(\frac{x^{2}}{2}+\frac{y^{2}}{1}=1\). Simplifying it, we can rewrite it as \(x^2 + 2y^2 = 2\).
2. The equation of the line is given by \(y = x + 1\).
3. Substitute \(y = x + 1\) in the ellipse equation:
   • \(x^2 + 2(x + 1)^2 = 2\)
   • Expanding: \(x^2 + 2(x^2 + 2x + 1) = 2\)
   • Further simplifying gives: \(x^2 + 2x^2 + 4x + 2 = 2\)
   • Then, \(3x^2 + 4x + 2 = 2\)
   • Simplify to: \(3x^2 + 4x = 0\)
   • Factorize to: \(x(3x + 4) = 0\)
   • Thus, \(x = 0\) or \(x = -\frac{4}{3}\)
4. Substitute back to find the corresponding \(y\)-values:
   • If \(x = 0\), then \(y = 1\).
   • If \(x = -\frac{4}{3}\), then \(y = -\frac{1}{3}\).
5. Points of intersection \(A\) and \(B\) are \((0, 1)\) and \(\left(-\frac{4}{3}, -\frac{1}{3}\right)\).
6. The center \(O\) of the ellipse is the origin \((0, 0)\).
7. Find the vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\):
   • \(\overrightarrow{OA} = (0 - 0, 1 - 0) = (0, 1)\)
   • \(\overrightarrow{OB} = \left(-\frac{4}{3} - 0, -\frac{1}{3} - 0\right) = \left(-\frac{4}{3}, -\frac{1}{3}\right)\)
8. Use the dot product to find the angle \(\theta\) between the vectors:
   • The dot product: \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0 \cdot \left(-\frac{4}{3}\right) + 1 \cdot \left(-\frac{1}{3}\right) = -\frac{1}{3}\)
   • The magnitudes: \(|\overrightarrow{OA}| = 1\), \( |\overrightarrow{OB}| = \sqrt{\left(-\frac{4}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \frac{\sqrt{17}}{3}\)
   • The angle formula: \(\cos \theta = \frac{-\frac{1}{3}}{1 \times \frac{\sqrt{17}}{3}}\)
   • Simplifying: \(\cos \theta = \frac{-1}{\sqrt{17}}\)
   • Hence, \(\theta = \cos^{-1}\left(\frac{-1}{\sqrt{17}}\right)\)
9.  
10. Convert \(\theta\) to an angle in the required form. Through calculation, it simplifies to: \(\angle AOB = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)\).

Therefore, the correct answer is \(\frac{\pi}{2} + \tan^{-1} \frac{1}{4}\).