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

Answer 1

The correct option is A $\frac{\pi }{2}+{\tan }^{-1}\left(\frac{1}{4}\right)$

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

Using homogenization
$x^2+2y^2=2(1{)}^2$
$\Rightarrow x^2+2y^2=2(x+y{)}^2$
$\Rightarrow x^2+2y^2=2x^2+2y^2+4xy$
$\Rightarrow x^2+4xy=0$
$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
$\Rightarrow \tan \theta =\left|\frac{2\sqrt{(2{)}^2-0}}{1+0}\right|$
$\Rightarrow \tan \theta =4$
So, required obtuse angle $=\pi -\theta$
$=\pi -{\tan }^{-1}4$
$=\pi -{\cot }^{-1}\left(\frac{1}{4}\right)$
$=\pi -(\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{4}\right))$
$=\frac{\pi }{2}+{\tan }^{-1}\left(\frac{1}{4}\right)$