Question

If the curve $x^2+2y^2 = 2$ intersects the line $x+y=1$ at two points P and Q, then the angle subtended by the line segment PQ at the origin is :

• A. $\frac{\pi}{2} + \tan^{-1}(\frac{1}{4})$
• B. $\frac{\pi}{2} - \tan^{-1}(\frac{1}{4})$
• C. $\frac{\pi}{2} + \tan^{-1}(\frac{1}{3})$
• D. $\frac{\pi}{2} - \tan^{-1}(\frac{1}{3})$

Hint:

The method of homogenization is a powerful tool to find the angle subtended by a chord at the origin. If the curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$ and the line is $lx+my=1$, the combined equation of the lines joining the origin to the points of intersection is found by making the curve's equation homogeneous.

Answer 1

The Correct Option is A

Homogenize the curve using \(x+y=1\): \[ x^2+2y^2=2(x+y)^2 \] \[ x^2+2y^2=2x^2+4xy+2y^2 \Rightarrow x^2+4xy=0 \] \[ x(x+4y)=0 \] The lines are: \[ x=0,\quad x+4y=0 \] Slopes: \[ m_1=\infty,\quad m_2=-\frac{1}{4} \] Angle between them: \[ \theta=\tan^{-1}(4) =\frac{\pi}{2}-\tan^{-1}\!\left(\frac{1}{4}\right) \] The obtuse angle is: \[ \pi-\theta=\frac{\pi}{2}+\tan^{-1}\!\left(\frac{1}{4}\right) \] Correct option: (A)