Question

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.

Answer 1

The equation of the family of parabolas with latus rectum $4a$ and axis parallel to the x-axis is given by
${\left(y-\beta \right)}^2=4a\left(x-\alpha \right)$ ...(1)
where $\alpha \mathrm{and}\beta$ are two arbitrary constants.
As this equation has two arbitrary constants, we shall get second order differential equation.
Differentiating equation (1) with respect to $x$, we get
$2\left(y-\beta \right)\frac{dy}{dx}=4a$ ...(2)
Differentiating equation (2) with respect to $x$, we get
$\left(y-\beta \right)\frac{d^{2}y}{d{x}^2}+\left(\frac{dy}{dx}\frac{dy}{dx}\right)=0$ ...(3)
Now, from equation (2), we get
$\left(y-\beta \right)=\frac{4a}{{\displaystyle \frac{dy}{dx}}}$ ...(4)
From (3) and (4), we get
$$\begin{gathered} \frac{2a}{{\displaystyle \frac{dy}{dx}}}\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0 \\ \Rightarrow 2a\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^3=0 \\ \mathrm{It}\mathrm{is}\text{the required differential equation.} \end{gathered}$$