JEE PYQ: Ellipse Question 3
Question 3 - 2021 (25 Feb Shift 2)
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) $\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$ (b) $\frac{\pi}{2} - \tan^{-1}\left(\frac{1}{4}\right)$ (c) $\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{3}\right)$ (d) $\frac{\pi}{2} - \tan^{-1}\left(\frac{1}{3}\right)$
Show Answer
Answer: (a) $\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$
Solution
Using homogenisation $x^2 + 2y^2 = 2(1)^2$
$x^2 + 2y^2 = 2(x + y)^2$
$x^2 + 2y^2 = 2x^2 + 2y^2 + 4xy$
$x^2 + 4xy = 0$
For $ax^2 + 2hxy + by^2 = 0$
$\tan \theta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right| = \left|\frac{2\sqrt{(2)^2 - 0}}{1 + 0}\right|$
$\tan \theta = -4$
$\theta = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$
BETA
