JEE Main 2022 — Mathematics Calculus

Given,

$$xdy=(\sqrt{{x}^{2}+{y}^{2}}+y)dx$$xdy=(x2+y2​+y)dx

$$\Rightarrow xdy-ydx=\sqrt{{x}^{2}+{y}^{2}}dx$$⇒xdy−ydx=x2+y2​dx

$$\Rightarrow \frac{xdy-ydx}{{x}^{2}}=\sqrt{1+\frac{{y}^{2}}{{x}^{2}}}\cdot \frac{dx}{x}$$⇒x2xdy−ydx​=1+x2y2​​⋅xdx​

$$\Rightarrow \frac{d(\frac{y}{x})}{\sqrt{1+{(\frac{y}{x})}^{2}}}=\frac{dx}{x}$$⇒1+(xy​)2​d(xy​)​=xdx​

Now integrating both side we get,

$$\Rightarrow \int \frac{d(\frac{y}{x})}{\sqrt{1+{(\frac{y}{x})}^{2}}}=\int \frac{dx}{x}$$⇒∫1+(xy​)2​d(xy​)​=∫xdx​

$$\Rightarrow \mathrm{ln}(\frac{y}{x}+\sqrt{{(\frac{y}{x})}^{2}+1})=\mathrm{ln}x+\mathrm{ln}c$$⇒ln(xy​+(xy​)2+1​)=lnx+lnc

$$\Rightarrow \frac{y+\sqrt{{y}^{2}+{x}^{2}}}{x}=cx$$⇒xy+y2+x2​​=cx

$$\Rightarrow y+\sqrt{{y}^{2}+{x}^{2}}=c{x}^{2}$$⇒y+y2+x2​=cx2

Now given when $x=1,y=0\Rightarrow 0+1=c\Rightarrow c=1$x=1,y=0⇒0+1=c⇒c=1

So equation of curve is $y+\sqrt{{x}^{2}+{y}^{2}}={x}^{2}$y+x2+y2​=x2

Now at $x=2,y=\alpha$x=2,y=α

So putting the value in curve we get, $2+\sqrt{4+{\alpha }^{2}}=4$2+4+α2​=4

$$\Rightarrow 4+{\alpha }^{2}=16+{\alpha }^{2}=8\alpha$$⇒4+α2=16+α2=8α

$$\Rightarrow \alpha =\frac{3}{2}$$⇒α=23​