Ellipse
Chord of Ellipse — Angle at Centre
nta_pyq_2026_jan
Grade 11
Question:
Let the line $y-x=1$ intersect the ellipse $\dfrac{x^2}{2}+y^2=1$ at the points $A$ and $B$. Then the angle made by the line segment $AB$ at the centre of the ellipse is:
$\dfrac{\pi}{2}+\tan^{-1}\!\left(\dfrac{1}{4}\right)$
$\dfrac{\pi}{2}+2\tan^{-1}\!\left(\dfrac{1}{4}\right)$
$\dfrac{\pi}{2}-\tan^{-1}\!\left(\dfrac{1}{4}\right)$
$\pi-\tan^{-1}\!\left(\dfrac{1}{4}\right)$
Step-by-Step Solution
Key Concept: Substitute $y=x+1$ into $\tfrac{x^2}{2}+y^2=1$. The intersections give $A=(0,1)$ and $B=\left(-\tfrac{4}{3},-\tfrac{1}{3}\right)$. Slope of $OB$: $m_{OB}=\tan\theta=\tfrac{1}{4}$.
$\angle AOB=\dfrac{\pi}{2}+\tan^{-1}\!\left(\dfrac{1}{4}\right)$.
Correct Answer: 1