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 \[\alpha\text{ and }\beta\]  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

\[\left( y - \beta \right)\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx}\frac{dy}{dx} \right) = 0............(4)\]
From (3) and (4), we get 
\[\frac{2a}{\frac{dy}{dx}}\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 = 0\]
\[ \Rightarrow 2a\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^3 = 0 \]
It is the required differential equation.