# If the curve

Question:

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

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

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

3. (3) $\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{3}\right)$

4. (4) $\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{3}\right)$

Correct Option: 1,

Solution:

Ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$

Line $x+y=1$

Using homogenisation $x^{2}+2 y^{2}=2(1)^{2}$

$x^{2}+2 y^{2}=2(x+y)^{2}$

$x^{2}+2 y^{2}=2 x^{2}+2 y^{2}+4 x y$

$x^{2}+4 x y=0$

for $a x^{2}+2 h x y+b y^{2}=0$

$\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|$

$\tan \theta=\left|\frac{2 \sqrt{(2)^{2}-0}}{1+0}\right|$

$\tan \theta=-4$

$\cot \theta=-\frac{1}{4}$

$\theta=\cot ^{-1}\left(-\frac{1}{4}\right)$

$\theta=\pi-\cot ^{-1}\left(\frac{1}{4}\right)$

$\theta=\pi-\left(\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{4}\right)\right)$

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