Question

Solve:

$\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$

• A. $\sqrt{x^2-y^2}+\sqrt{1+x^2-y^2}=c\left(\frac{x-y}{\sqrt{x^2-y^2}}\right)$
• B. $\sqrt{x^2-y^2}+\sqrt{1+x^2+y^2}=c\left(\frac{x-y}{\sqrt{x^2+y^2}}\right)$
• C. $\sqrt{x^2-y^2}+\sqrt{1+x^2-y^2}=c\left(\frac{x+y}{\sqrt{x^2-y^2}}\right)$
• D. $\sqrt{x^2+y^2}+\sqrt{1+x^2-y^2}=c\left(\frac{x+y}{\sqrt{x^2+y^2}}\right)$

Answer 1

The correct option is C $\sqrt{x^2-y^2}+\sqrt{1+x^2-y^2}=c\left(\frac{x+y}{\sqrt{x^2-y^2}}\right)$

$\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$

Substitute $x=r\sec \theta$, $y=r\tan \theta$

$\therefore x^2-y^2=r^2$ ..... $\left(1\right)$

and $\sin \theta =\frac{y}{x}$ ..... $\left(2\right)$

On differentiating $\left(1\right)$, $xdx-ydy=rdr$... $\left(3\right)$

On differentiating $\left(2\right)$,

$xdy-ydx=x^2\cos \theta d\theta =r^2\sec \theta d\theta$ ..... $\left(4\right)$

Now $\frac{rdr}{r^2\sec \theta d\theta }=\frac{\sqrt{1+r^2}}{r}$

$\Rightarrow \int \frac{dr}{\sqrt{1+r^2}}=\int \sec \theta d\theta$

$\Rightarrow \ln (r+\sqrt{1+r^2})=\ln (\sec \theta +\tan \theta )+inc$

$\Rightarrow r+\sqrt{1+r^2}=c(\sec \theta +\tan \theta )$

$\sqrt{x^2-y^2}+\sqrt{1+x^2-y^2}=c\left(\frac{x+y}{\sqrt{x^2-y^2}}\right)$