Coordinate Geometry – Ellipse – JEE Main 23 Jan 2026 Shift 1

Solution:

Equation of ellipse: $x^2 + 2y^2 = 2$.
Equation of line: $y = x+1$.

Substitute $y$ in the ellipse equation to find intersection points:
$$ x^2 + 2(x+1)^2 = 2 $$
$$ x^2 + 2(x^2 + 2x + 1) = 2 $$
$$ 3x^2 + 4x = 0 $$

Roots are $x = 0$ and $x = -4/3$.

If $x = 0$, $y = 1$. Point $A = (0, 1)$.
If $x = -4/3$, $y = -4/3 + 1 = -1/3$. Point $B = (-4/3, -1/3)$.


We need the angle $\angle AOB$ where $O$ is the origin $(0,0)$.
Slope of $OA$, $m_1 = \frac{1-0}{0-0} = \infty$ (Vertical line along y-axis).
Slope of $OB$, $m_2 = \frac{-1/3 – 0}{-4/3 – 0} = \frac{1}{4}$.

Let $\theta$ be the angle between $OA$ and $OB$.
Since $OA$ lies on the positive y-axis (angle $\pi/2$) and $OB$ lies in the 3rd quadrant.

Let $\alpha$ be the angle $OB$ makes with the positive x-axis. $\tan \alpha = 1/4$.
Since $B$ is in the 3rd quadrant, the actual angle of $OB$ is $\pi + \tan^{-1}(1/4)$.
The angle of $OA$ is $\pi/2$.

The angle between them is:
$$ \theta = (\pi + \tan^{-1}(1/4)) – \frac{\pi}{2} $$
$$ \theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right) $$

Ans. (2)