The solution of (1+y^2) dx- xydy=0 ,y (1 )=0 represents a conic. Its eccentricity is

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Correct answer: D. (d) Given (1+y^2) dx- xydy=0 and y (1 )=0. Take, (1+y^2) dx- xydy=0 ⇒ dx(1+y^2)= xydy dxx= ydy1+y^2 Let 1+y^2=t ⇒ ydy= 12 dt log ⁡x= 12 dtt ⇒ log ⁡x= 12 log ⁡t+ log ⁡c log ⁡x= log ⁡ ( tc ) ⇒x= 1+y^2c By putting x=1 ,y=0, we get c=1 Then, x= 1+y^2 ⇒x^2-y^2=1 a=b=1c= 1+1= 2 So , e= ca= 21= 2 .

$The solution of (1 + y^{2}) dx - xydy = 0, y (1) = 0 represents a conic. Its eccentricity is$

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