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.
Show 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.
To find the angle subtended by a chord at the origin, we use the Method of Homogenization. This technique combines the equation of the curve and the line to create a joint equation for the two lines ($OA$ and $OB$) connecting the origin to the intersection points.
Step 1: Align the equations.
The given ellipse is:
$$\frac{x^2}{2} + y^2 = 1 \quad \dots (i)$$
The given line is:
$$y - x = 1 \quad \dots (ii)$$
Step 2: Create the homogeneous equation.
Since the right-hand side of the line equation is already $1$, we can substitute it into the constant term of the ellipse equation to make it a second-degree homogeneous equation:
$$\frac{x^2}{2} + y^2 = (1)^2$$
$$\frac{x^2}{2} + y^2 = (y - x)^2$$
Step 3: Simplify the pair of straight lines.
Expanding the right side:
$$\frac{x^2}{2} + y^2 = y^2 - 2xy + x^2$$
Subtracting $y^2$ from both sides and gathering all terms on one side:
$$x^2 - \frac{x^2}{2} - 2xy = 0$$
$$\frac{x^2}{2} - 2xy = 0$$
Multiply by $2$ to simplify:
$$x^2 - 4xy = 0 \implies x(x - 4y) = 0$$
Step 4: Calculate the angle.
The resulting equation represents two distinct lines:
Line 1: $x = 0$ (This is the $y$-axis, which is vertical).
Line 2: $y = \frac{1}{4}x$ (This line has a slope $m = \frac{1}{4}$, so its angle with the $x$-axis is $\tan^{-1}(\frac{1}{4})$).
Because the first line is the $y$-axis ($\frac{\pi}{2}$ from the $x$-axis), the angle $\theta$ between them is the complement of the second line's angle:
$$\theta = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{4}\right)$$
Conclusion:
The angle subtended by segment $AB$ at the center is:
