The differential equation corresponding to th | vedclass

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(A) Given the family of parabolas is $y^2 = 4a(x+a) \quad ...(i)$Differentiating with respect to $x$,we get:$2y \frac{dy}{dx} = 4a$$\Rightarrow a = \f

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$y\left(\frac{dy}{dx}\right)^2+2x\frac{dy}{dx}-y=0$

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(A) Given the family of parabolas is $y^2 = 4a(x+a) \quad ...(i)$Differentiating with respect to $x$,we get:$2y \frac{dy}{dx} = 4a$$\Rightarrow a = \frac{1}{2} y \frac{dy}{dx} \quad ...(ii)$Substituting the value of $a$ from equation $(ii)$ into equation $(i)$:$y^2 = 4 \left( \frac{1}{2} y \frac{dy}{dx} ight) \left( x + \frac{1}{2} y \frac{dy}{dx} ight)$$y^2 = 2y \frac{dy}{dx} \left( x + \frac{1}{2} y \frac{dy}{dx} ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} ight)^2$Rearranging the terms,we get:$y \left( \frac{dy}{dx} ight)^2 + 2x \frac{dy}{dx} - y = 0$

The differential equation corresponding to the family of parabolas $y^2=4a(x+a)$,where $a$ is the parameter,is

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