A) order 1, degree 2
B) order 1, degree 1
C) order 1, degree 3
D) order 2, degree 2
Correct Answer: C
Solution :
Given, equation of family of curve is \[{{y}^{2}}=2c(x+\sqrt{c})\] ...(i) \[2y.\frac{dy}{dx}=2c\] \[\Rightarrow \] \[c=y.{{y}_{1}}\] ...(ii) From Eq. (i), \[{{y}^{2}}=2y{{y}_{1}}(x+\sqrt{y{{y}_{1}}})\] \[\Rightarrow \] \[{{y}^{2}}=2y{{y}_{1}}x+2{{(y{{y}_{1}})}^{3/2}}\] \[\Rightarrow \] \[{{({{y}^{2}}-2y{{y}_{1}}x)}^{2}}=4{{(y{{y}_{1}})}^{3}}\] Hence, order = 1 and degree = 3.
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