JEE Mathematics 2026 Question with Solution

Given: The line is $y - x = 1$ and the ellipse is $\frac{x^2}{2} + \frac{y^2}{1} = 1$.

Find: The angle subtended by chord AB at the center $\left(0,0\right)$.

Using the ellipse in simplified form,

$$x^2 + 2y^2 = 2$$

Homogenize it with the line $y - x = 1$ by replacing $1$ with $y-x$:

$$x^2 + 2y^2 = 2(y-x)^2$$

Now expand:

$$x^2 + 2y^2 = 2(y^2 + x^2 - 2xy)$$ $$x^2 + 2y^2 = 2y^2 + 2x^2 - 4xy$$

Rearranging,

$$x^2 - 4xy = 0$$ $$x(x - 4y) = 0$$

So the pair of lines through the origin joining the center to the points of intersection are:

$$x = 0$$

and

$$y = \frac{1}{4}x$$

The line $x = 0$ is the Y-axis. The line $y = \frac{1}{4}x$ makes an angle $\alpha = \tan^{-1}\left(\frac{1}{4}\right)$ with the positive X-axis.

Therefore, the angle between the Y-axis and this line is

$$\theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$$

Hence, the correct option is C.

Given: A chord of the ellipse is cut by the line $y-x=1$.

Find: The angle made by that chord at the center.

The idea is to obtain the combined equation of the two lines from the origin to the intersection points A and B. For this, write the ellipse as

$$x^2 + 2y^2 = 2(1)^2$$

Since the given line gives $1 = y-x$, substitute this into the right-hand side during homogenization:

$$x^2 + 2y^2 = 2(y-x)^2$$

This produces the joint equation of the pair of lines through the origin and the points A and B.

After simplification,

$$x^2 + 2y^2 = 2y^2 + 2x^2 - 4xy$$ $$x^2 - 4xy = 0$$ $$x(x-4y)=0$$

Thus the two lines are $x=0$ and $y=\frac{x}{4}$.

Now measure the angle between them at the origin. One line is at angle $\frac{\pi}{2}$ and the other at angle $\tan^{-1}\left(\frac{1}{4}\right)$. From the geometry used in the solution, the required angle is

$$\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$$

So the correct option is C.