A) \[yy'-2xyy'+{{y}^{2}}=0\]
B) \[yy'(yy'+2x)+{{y}^{2}}=0\]
C) \[yy'(yy'-2x)+{{y}^{2}}=0\]
D) \[yy'-2xyy'+y=0\]
Correct Answer: C
Solution :
[c] Given curve is \[{{y}^{2}}=4a(x-a)\] ? (i) On differentiating w.r.t. x we get \[2yy'=4a\] \[\Rightarrow a=\frac{yy'}{2}\] O putting the value of a in Eq. (i) we get \[{{y}^{2}}=4\left( \frac{yy'}{2} \right)\left( x-\frac{yy'}{2} \right)=yy'(2x-yy')\] \[\Rightarrow yy'(yy'-2x)+{{y}^{2}}=0\]
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