Question

The degree of the differential equation satisfying the relation

$\sqrt{1+x^2}+\sqrt{1+y^2}=\lambda (x\sqrt{1+y^2}-y\sqrt{1+x^2})$ is

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

The correct option is A

1

On putting $x=\text{tan A}$, $y=\text{tan B}$ we get

$\text{sec A}+\text{sec B}=\lambda \left(\text{tan A sec B – tan B sec A}\right)$

$\text{cos A + cos B}=\lambda \left(\text{sin A – sin B}\right)$

$\text{tan}\left(\frac{A-B}{2}\right)=\frac{1}{\lambda }$

${\text{tan}}^{-1}x-{\text{tan}}^{-1}y=2ta{n}^{-1}\frac{1}{\lambda }$

On differentiating $\frac{1}{1+x^2}-\frac{1}{1+y^2}\frac{dy}{dx}=0$