A) \[2({{x}^{2}}-{{y}^{2}})y'=xy\]
B) \[2({{x}^{2}}+{{y}^{2}})y'=xy\]
C) \[({{x}^{2}}-{{y}^{2}})y'=2xy\]
D) \[({{x}^{2}}+{{y}^{2}})y'=2xy\]
Correct Answer: C
Solution :
Given equation of family of curves is \[{{x}^{2}}+{{y}^{2}}-2ay=0\] ...(i) On differentiating w.r.t.\[x,\]we get \[2x+2yy'-2ay'=0\] \[\Rightarrow \] \[2x+2yy'=2ay'\] \[\Rightarrow \] \[\frac{2x+2yy'}{y'}=2a\] ?..(ii) From Eq. (i), we get \[2a=\frac{{{x}^{2}}+{{y}^{2}}}{y}\] On putting this value of 2a in Eq. (ii), we get \[\frac{2x+2yy'}{y'}=\frac{{{x}^{2}}+{{y}^{2}}}{y}\] \[\Rightarrow \] \[2xy+2{{y}^{2}}y'={{x}^{2}}y+{{y}^{2}}y'\] \[\Rightarrow \] \[({{x}^{2}}-{{y}^{2}})y'=2xy\]
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