A) order 1, degree 3
B) order 2, degree 2
C) degree 3, order 3
D) degree 4, order 4
Correct Answer: A
Solution :
Given family of curves \[{{y}^{2}}=2c(x+\sqrt{c})\] ?(i) On differentiating both sides, we get \[2y\frac{dy}{dx}=2c(1+0)\Rightarrow c=y\frac{dy}{dx}\] From Eq. (i), \[{{y}^{2}}=2y\frac{dy}{dx}\left\{ x+{{\left( y\frac{dy}{dx} \right)}^{1/2}} \right\}\] \[\Rightarrow \] \[\left( {{y}^{2}}-2xy\frac{dy}{dx} \right)=2{{\left( y\frac{dy}{dx} \right)}^{3/2}}\] On squaring both sides, we get \[{{\left( {{y}^{2}}-2xy\frac{dy}{dx} \right)}^{2}}=4{{\left( y\frac{dy}{dx} \right)}^{3}}\] So, order 1 and degree 3.
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