Question

Let the line \( y - x = 1 \) intersect the ellipse \( \frac{x^2}{2} + \frac{y^2}{1} = 1 \) at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is:

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

Hint:

To homogenize, ensure the line equation is in the form of "Expression = 1". Substitute this "1" into the constant or second-degree terms of the curve to make all terms degree 2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the angle subtended by a chord (the line segment AB) at the center of the ellipse $(0,0)$, we use the method of homogenization. This process creates a joint equation for the two lines connecting the origin to the intersection points A and B.
Step 2: Key Formula or Approach:
1. Equation of Line: \( y - x = 1 \).
2. Equation of Ellipse: \( x^2 + 2y^2 = 2 \).
3. Homogenize the ellipse equation using the line equation: \( \frac{y-x}{1} = 1 \).
Step 3: Detailed Explanation:
Homogenizing the ellipse \( x^2 + 2y^2 = 2(1)^2 \) using the line: \[ x^2 + 2y^2 = 2(y - x)^2 \] \[ x^2 + 2y^2 = 2(y^2 + x^2 - 2xy) \] \[ x^2 + 2y^2 = 2y^2 + 2x^2 - 4xy \] Rearranging terms to one side: \[ x^2 - 4xy = 0 \implies x(x - 4y) = 0 \] This represents two lines passing through the origin: Line 1: \( x = 0 \) (the Y-axis). Line 2: \( y = \frac{1}{4}x \) (a line with slope \( m = 1/4 \)). The angle \( \alpha \) that the line \( y = \frac{1}{4}x \) makes with the X-axis is \( \tan^{-1}(1/4) \). Since the Y-axis is perpendicular to the X-axis, the angle between them is \( 90^\circ \). The total angle \( \theta \) between the Y-axis and the line in the second quadrant is: \[ \theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right) \] Step 4: Final Answer:
(3) \( \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right) \).