question_answer

If the differential equation representing the family of all circles touching x-axis at the origin is \[\left( {{x}^{2}}-{{y}^{2}} \right)\frac{dy}{dx}=g\left( x \right)y,\]then g(x) equals:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

A) \[\frac{1}{2}x\]                                

B) \[2{{x}^{2}}\]

C) \[2x\]                                   

D) \[\frac{1}{2}{{x}^{2}}\]

Correct Answer: C

Solution :

Since family of all circles touching x-axis at the origin \[\therefore \]Eqn is \[{{(x)}^{2}}+{{(y-a)}^{2}}={{a}^{2}}\]where (0, a) is the centre of circle. \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+{{a}^{2}}-2ay={{a}^{2}}\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-2ay=0\]                                     ...(1) Differentiate both side w.r.t 'x', we get \[2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0\] \[\Rightarrow \]\[x+y\frac{dy}{dx}-a\frac{dy}{dx}=0\Rightarrow \frac{x+y\frac{dy}{dx}}{dy}=a\]\[\Rightarrow \] Put value of 'a' in eqn (1), we get \[{{x}^{2}}+{{y}^{2}}-2y\left[ \frac{y\frac{dy}{dx}+x}{\frac{dy}{dx}} \right]=0\] \[\Rightarrow \]\[({{x}^{2}}+{{y}^{2}})\frac{dy}{dx}-2{{y}^{2}}\frac{dy}{dx}-2xy=0\] \[\Rightarrow \]\[({{x}^{2}}+{{y}^{2}}-2{{y}^{2}})\frac{dy}{dx}=2xy\] \[\Rightarrow \]\[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=2xy\equiv g(x)y\] Hence, g(x) = 2x