JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

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
$π - tan^{- 1} (\frac{1}{4})$
B
$\frac{π}{2} + tan^{- 1} (\frac{1}{4})$
C
$\frac{π}{2} + 2 tan^{- 1} (\frac{1}{4})$
D
$\frac{π}{2} - tan^{- 1} (\frac{1}{4})$

Solution

Find the intersection points. Substitute $y = x + 1$ into the ellipse:

$\frac{x ^{2}}{2} + (x + 1)^{2} = 1 \Rightarrow \frac{3 x ^{2}}{2} + 2 x = 0 \Rightarrow x (\frac{3 x}{2} + 2) = 0$

So $x = 0$ giving $A = (0, 1)$ , or $x = - \frac{4}{3}$ giving $B = (- \frac{4}{3}, - \frac{1}{3})$ .

Angles at the centre (origin). OA lies along the positive $y$ -axis, so $∠$ OA $= \frac{π}{2}$ from the $x$ -axis.

For OB: slope $= \frac{- 1/3}{- 4/3} = \frac{1}{4}$ , but B lies in the third quadrant, so OB makes an angle $π + tan^{- 1} \frac{1}{4}$ .

$∠$ AOB $= (π + tan^{- 1} \frac{1}{4}) - \frac{π}{2} = \frac{π}{2} + tan^{- 1} \frac{1}{4}$ .

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