Question

The curve satisfying $y_1=\frac{y^2-2xy-x^2}{y^2+2xy-x^2}$ and passing through $(1,-1)$ is:

• A. a straight line
• B. a circle
• C. an. ellipse
• D. none of these

Answer 1

Answer: (A)

a straight line

$y_1=\frac{y^2-2xy-x^2}{y^2+2xy-x^2}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow \frac{dx}{x}=\left(\frac{v^2+2v-1}{v^3+v^2+v+1}\right)dv$

$=-\left(\frac{v^2+2v-1}{(v+1)(v^2+1)}\right)dv=(\frac{-1}{(v+1)}+\frac{2v}{(v^2+1)})dv$

Integrating both sides to get

$\log x-\log (v+1)+\log (v^2+1)+\log c=0$

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

$\Rightarrow c(y^2+x^2)=y+x$

If this curve passes through $(1,-1)$, we get $2c=0\Rightarrow c=0$

Hence the required curve is $y+x=0$

which is a straight line.