A) a circle with centre on the y-axis
B) an ellipse with major axis along the y-axis
C) a circle with centre on the x-axis
D) a hyperbola with transverse axis along the x-axis
Correct Answer: C
Solution :
\[({{x}^{2}}-{{y}^{2}})dx+2xydy=0\] \[\frac{dy}{dx}\,=\,\frac{-({{x}^{2}}-{{y}^{2}})}{2xy}\] Put \[y=vx\] \[\frac{dy}{dx}\,\,=\,\,v+\frac{xdv}{dx}\] \[\Rightarrow \,\,v+\frac{xdv}{dx}\,=\,\frac{-({{x}^{2}}-{{v}^{2}}{{x}^{2}})}{2x(vx)}\] \[\Rightarrow \,\,\,\,v+\frac{xdv}{dx}\,=\,\frac{-1+{{v}^{2}}}{2\,v}\] \[\Rightarrow \,\,\,\,x\frac{dv}{dx}\,=\,\frac{-1+{{v}^{2}}}{2\,v}-v\] \[\Rightarrow \,\,\,\,x\frac{dv}{dx}\,=\,\frac{-1-{{v}^{2}}}{2\,v}\] \[\Rightarrow \,\,\,\int{\frac{2vdv}{1+{{v}^{2}}}\,\,=\,\,-\int{\frac{dy}{x}}}\] \[\Rightarrow \,\,\,\text{ }log\,\,(1+{{v}^{2}})=-\,log\,x+log\,C\] \[\Rightarrow \,\,\,\,\,1+\frac{{{y}^{2}}}{{{x}^{2}}}\,\,=\,\,\frac{C}{x}\] \[\Rightarrow \,\,\,\,\,{{x}^{2}}+{{y}^{2}}=Cx\] It passes through (1, 1) \[\therefore \,\,\,\,\,\,C=2\] Curve \[{{x}^{2}}+{{y}^{2}}-2x=0\]
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