A) 2
B) -2
C) 3
D) -4
Correct Answer: B
Solution :
We have,\[\frac{dy}{dx}=\frac{ax+3}{2y+f}\] \[\Rightarrow \] \[(ax+3)dx=(2y+f)dy\] On integrating, we obtain \[a\frac{{{x}^{2}}}{2}+3y={{y}^{2}}+fy+c\] \[\Rightarrow \] \[-\frac{a}{2}{{x}^{2}}+{{y}^{2}}-3x+fy+c=0\] This will represent a circle, if \[-\frac{a}{2}=1\] (\[\because \]coefficient of\[{{x}^{2}}=\]coefficient of\[{{y}^{2}}\]) \[\Rightarrow \] \[a=-2\]You need to login to perform this action.
You will be redirected in
3 sec