The differential equation of the family of pa | vedclass

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(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is $(y-0)^2 = 4a(x+a)$,where $a$ is a parameter. This simpli

Vedclass · Accepted Answer

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

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(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is $(y-0)^2 = 4a(x+a)$,where $a$ is a parameter. This simplifies to $y^2 = 4ax + 4a^2$. Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2} \frac{dy}{dx}$. Substituting this value of $a$ into the original equation: $y^2 = 4\left(\frac{y}{2} \frac{dy}{dx}ight)x + 4\left(\frac{y}{2} \frac{dy}{dx}ight)^2$. $y^2 = 2xy \frac{dy}{dx} + 4\left(\frac{y^2}{4} \left(\frac{dy}{dx}ight)^2ight)$. $y^2 = 2xy \frac{dy}{dx} + y^2 \left(\frac{dy}{dx}ight)^2$. Dividing by $y$ (assuming $y 
eq 0$): $y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}ight)^2$. Rearranging the terms gives $-y \left(\frac{dy}{dx}ight)^2 = 2x \frac{dy}{dx} - y$.

The differential equation of the family of parabolas with the focus at the origin and the axis along the $X$-axis is:

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