The differential equation of the family of pa | vedclass

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Question

(B) The equation of a family of parabolas with focus at $(0, 0)$ and the $x$-axis as the axis is given by $y^2 = 4a(x + a)$,where $a$ is a parameter.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of a family of parabolas with focus at $(0, 0)$ and the $x$-axis as the axis is given by $y^2 = 4a(x + a)$,where $a$ is a parameter. Differentiating with respect to $x$: $2y \frac{dy}{dx} = 4a$ $\Rightarrow a = \frac{y}{2} \frac{dy}{dx}$. Substituting the value of $a$ back into the original equation: $y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} ight) \left( x + \frac{y}{2} \frac{dy}{dx} ight)$ $y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} ight)$ Dividing by $y$ (assuming $y 
eq 0$): $y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} ight)^2$ Rearranging the terms: $-y \left( \frac{dy}{dx} ight)^2 = 2x \frac{dy}{dx} - y$. Thus,the correct option is $B$.

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

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