A) a hyperbola
B) a parabola
C) an ellipse
D) a circle
Correct Answer: A
Solution :
\[\frac{dy}{dx}=\frac{{{x}^{2}}+{{y}^{2}}+1}{2xy}\] \[\Rightarrow \] \[2xy\,dy=({{x}^{2}}+1)dx+{{y}^{2}}dx\] \[\Rightarrow \] \[xd({{y}^{2}})-{{y}^{2}}dx=({{x}^{2}}+1)dx\] \[\Rightarrow \] \[\frac{xd({{y}^{2}})-{{y}^{2}}dx}{{{x}^{2}}}=\left( 1+\frac{1}{{{x}^{2}}} \right)dx\] \[\Rightarrow \] \[\frac{{{y}^{2}}}{x}=x-\frac{1}{x}+c\] \[\Rightarrow \] \[{{y}^{2}}={{x}^{2}}-1+cx\] At \[x=1,\text{ }y=0,\text{ }\Rightarrow c=0\] \[\therefore \] \[{{y}^{2}}={{x}^{2}}-1\] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}=1\] which is a hyperbola.
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