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(D) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as the axis of symmetry is given by $y^2 = 4a(x+a)$,wher

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$1$

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(D) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as the axis of symmetry is given by $y^2 = 4a(x+a)$,where $a$ is an arbitrary parameter.Expanding this,we have $y^2 = 4ax + 4a^2$.Differentiating with respect to $x$,we get:$2y \frac{dy}{dx} = 4a \Rightarrow a = \frac{y}{2} \frac{dy}{dx}$.Substituting the value of $a$ back into the original equation $y^2 = 4a(x+a)$:$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} ight) \left( x + \frac{y}{2} \frac{dy}{dx} ight)$$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} ight)^2$$y \left( \frac{dy}{dx} ight)^2 + 2x \frac{dy}{dx} - y = 0$.Since the equation involves only the first derivative $\frac{dy}{dx}$,the order of the differential equation is $1$.

The order of the differential equation corresponding to the family of parabolas whose axes are along the $X$-axis and whose foci are at the origin,is

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