Question

The curve passes through $(1,1)$ satisfies the differential equation
$\frac{dy}{dx}+\sqrt{(1-\frac{1}{x^2})(1-\frac{1}{y^2})}=0$
If it passes through $(\sqrt{2},k)$ then the possible value(s) of $\left[k\right]$ is/are
(where [.] denote the greatest integer function)

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (A), (D)

$-2$

$\frac{dy}{dx}+\sqrt{(1-\frac{1}{x^2})(1-\frac{1}{y^2})}=0$
$\Rightarrow \frac{dy}{\sqrt{(1-\frac{1}{y^2})}}=-\sqrt{(1-\frac{1}{x^2})}dx$

Let $y=\sec \theta$ and $x=\sec \varphi$
$dy=\sec \theta \tan \theta d\theta ,dx=\sec \varphi \tan \varphi d\varphi$

$\Rightarrow {\sec }^2\theta d\theta =-({\sec }^2\varphi -1)d\varphi$
Integrating both the sides
$\tan \theta =-\tan \varphi +\varphi +C\Rightarrow \sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}x+C$
It passes through $(1,1)$ so,
$\Rightarrow C=0$

Putting $(\sqrt{2},k)$
$\Rightarrow \sqrt{k^2-1}=-1+\frac{\pi }{4}\Rightarrow k^2={(\frac{\pi }{4}-1)}^2+1\Rightarrow 1<k^2<2[\because 0<{(\frac{\pi }{4}-1)}^2<1]\Rightarrow \left[k\right]=1,-2$