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JEE Main 202623 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Ellipse

Let the line y-x=1 intersect the ellipse x^ 2 2 + y^ 2 1 =1 at the...

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 : Let the line y-x=1 intersect the ellipse x^ 2 2 + y^ 2 1 =1 at the points A and B. Then the angle made by the line

Let the line y-x=1 intersect the ellipse x^ 2 2 + y^ 2 1 =1 at the points A and B. Then the angle made by the line

Options

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

Correct answer

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

Step-by-step solution

Let the line y-x=1 intersect the ellipse x^ 2 2 + y^ 2 1 =1 at the points A and B. Then the angle made by the line

By solving line & equation of ellipse we get

x = 0

&

x = -\dfrac{4}{3}

\therefore B\left(-\dfrac{4}{3}, -\dfrac{1}{3}\right)

m_{OB} = \tan\theta = \dfrac{1}{4}

\therefore \angle AOB = \dfrac{\pi}{2} + \theta = \dfrac{\pi}{2} + \tan^{-1}\dfrac{1}{4}

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