Let the line y-x=1 intersect the ellipse x^22+y^2=1 at the points A and B.

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:

Accepted Answer

Substituting $y=x+1$ into $\dfrac{x^2}{2}+y^2=1$:

$$\frac{x^2}{2}+x^2+2x+1=1 \;\Rightarrow\; \frac{3x^2}{2}+2x=0 \;\Rightarrow\; x=0,\ -\frac43.$$

So $A=(0,1)$, $B=\left(-\dfrac43,-\dfrac13\right)$.

$OA$ makes angle $90^\circ$ with the $x$-axis. For $OB$, $\tan\theta = \dfrac{-1/3}{-4/3}=\dfrac14$, and $B$ lies in the third quadrant, so $OB$ makes angle $180^\circ+\tan^{-1}\left(\dfrac14\right)$.

$$\angle AOB = 180^\circ+\tan^{-1}\left(\tfrac14\right)-90^\circ = \frac{\pi}{2}+\tan^{-1}\left(\frac14\right).$$

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