The differential equation of the family of pa | vedclass

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Question

(A) The equation of a parabola with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = 4a(x+a)$,which simplifies to $y^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a parabola with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = 4a(x+a)$,which simplifies to $y^2 = 4a(x+a)$.Differentiating both sides with respect to $x$,we get $2y\frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2}\frac{dy}{dx}$.Substituting the value of $a$ into the original equation $y^2 = 4a(x+a)$,we get:$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 $y 
eq 0$):$y = 2x\frac{dy}{dx} + y\left(\frac{dy}{dx}ight)^2$Rearranging the terms,we get $y\left(\frac{dy}{dx}ight)^2 + 2x\frac{dy}{dx} - y = 0$,which is equivalent to $y\left(\frac{dy}{dx}ight)^2 = y - 2x\frac{dy}{dx}$ or $-y\left(\frac{dy}{dx}ight)^2 = 2x\frac{dy}{dx} - y$.

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

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