The differential equation of all parabolas ea | vedclass

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Question

(C) The equation of the family of parabolas with latus rectum $4a$ and axis parallel to the $x$-axis is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ are arb

Vedclass · Accepted Answer

of order $2$ and degree $1$

Vedclass · Answer

(C) The equation of the family of parabolas with latus rectum $4a$ and axis parallel to the $x$-axis is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ are arbitrary constants.Differentiating with respect to $x$:$2(y - k) \frac{dy}{dx} = 4a$$\Rightarrow (y - k) \frac{dy}{dx} = 2a$Differentiating again with respect to $x$:$(y - k) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$From the first derivative,we have $(y - k) = \frac{2a}{dy/dx}$. Substituting this into the second derivative equation:$\left( \frac{2a}{dy/dx} \right) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$Multiplying by $\frac{dy}{dx}$:$2a \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 = 0$In this differential equation,the highest order derivative is $\frac{d^2y}{dx^2}$,so the order is $2$. The power of the highest order derivative is $1$,so the degree is $1$.

The differential equation of all parabolas each of which has a latus rectum $4a$ and whose axes are parallel to the $x$-axis is:

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