Question

The line \(y = x + 1\) intersects the ellipse \[ \frac{x^2}{2} + \frac{y^2}{1} = 1 \] at points \(A\) and \(B\). Find the angle subtended by the segment \(AB\) at the centre of the ellipse.

• A. \( \dfrac{\pi}{2} + \tan^{-1}\!\left(\dfrac{1}{4}\right) \)
• B. \( \dfrac{\pi}{2} - \tan^{-1}\!\left(\dfrac{1}{4}\right) \)
• C. \( \dfrac{\pi}{2} + 2\tan^{-1}\!\left(\dfrac{1}{4}\right) \)
• D. \( \dfrac{\pi}{4} + \tan^{-1}\!\left(\dfrac{1}{4}\right) \)

Hint:

When one of the lines is vertical, subtract the inclination of the other line from \( \frac{\pi}{2} \) to get the angle between them.

Answer 1

The Correct Option is B

Concept:

The centre of the ellipse is at the origin \(O(0,0)\).
The angle subtended by chord \(AB\) at the centre is the angle between vectors \(\vec{OA}\) and \(\vec{OB}\).
If slopes of lines \(OA\) and \(OB\) are \(m_1\) and \(m_2\), then \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
Step 1: Find the points of intersection. Substitute \(y = x + 1\) into the ellipse: \[ \frac{x^2}{2} + (x+1)^2 = 1 \] \[ \frac{x^2}{2} + x^2 + 2x + 1 = 1 \] \[ \frac{3x^2}{2} + 2x = 0 \] \[ x(3x + 4) = 0 \] \[ x = 0,\quad x = -\frac{4}{3} \] Corresponding points: \[ A(0,1),\quad B\!\left(-\frac{4}{3},-\frac{1}{3}\right) \]
Step 2: Slopes of \(OA\) and \(OB\). \[ m_1 = \frac{1}{0} \Rightarrow \text{vertical line} \] \[ m_2 = \frac{-1/3}{-4/3} = \frac{1}{4} \]
Step 3: Angle between the two lines. Angle between vertical line and a line of slope \(m\): \[ \theta = \frac{\pi}{2} - \tan^{-1}(m) \] \[ \theta = \frac{\pi}{2} - \tan^{-1}\!\left(\frac{1}{4}\right) \]