A) order 1 and degree 3
B) order 1 and degree 2
C) order 2 and degree 1
D) order 3 and degree 1
Correct Answer: A
Solution :
Given curve is \[{{y}^{2}}=2c(x+\sqrt{c})\] ?.(i) \[\therefore \] \[2y\,\,\frac{dy}{dx}=2c\] \[\Rightarrow \] \[c=y\,\frac{dy}{dx}\] On putting the value of c in Eq. (i), we get \[{{y}^{2}}=2y\frac{dy}{dx}\left[ x+{{\left( y\frac{dy}{dx} \right)}^{1/2}} \right]\] \[\Rightarrow \] \[{{y}^{2}}-2yx\frac{dy}{dx}=2{{\left( y\frac{dy}{dx} \right)}^{3/2}}\] On squaring both sides, we get \[{{\left[ {{y}^{2}}-2xy\,\frac{dy}{dx} \right]}^{2}}=4{{\left[ y\frac{dy}{dx} \right]}^{3}}\] \[\therefore \] order =1, degree =3
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