JEE Main 2024 — 27 January, Shift 2 — Question 26

$\frac{d y}{d x}=\frac{x+y-2}{x-y}$

$x=X+h, y=Y+k$

$\frac{d Y}{d X}=\frac{X+Y}{X-Y}$

$\left.\begin{array}{l}h+k-2=0, h-k=0\end{array}\right\}$

Solving and we get :

$h=k=1$

$\mathrm{Y}=\mathrm{v} \mathrm{X}$

$v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^{2}}{1-v}$

$\frac{1-v}{1+v^{2}} d v=\frac{d X}{X}$

$\tan ^{-1} \mathrm{v}-\frac{1}{2} \ln \left(1+\mathrm{v}^{2}\right)=\ln |\mathrm{X}|+\mathrm{C}$

As curve is passing through $(2,1)$ $\tan ^{-1}\left(\frac{\mathrm{y}-1}{\mathrm{x}-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{\mathrm{y}-1}{\mathrm{x}-1}\right)^{2}\right)=\ln |\mathrm{x}-1|$

$\therefore \alpha=1$ and $\beta=2$

$\Rightarrow 5 \beta+\alpha=11$