Question

The differential equation satisfied by the system of parabolas $y^2=4a(x+a)$ is

• A. $y{\left(\frac{dy}{dx}\right)}^2-2x\left(\frac{dy}{dx}\right)-y=0$
• B. $y{\left(\frac{dy}{dx}\right)}^2-2x\left(\frac{dy}{dx}\right)+y=0$
• C. $y{\left(\frac{dy}{dx}\right)}^2+2x\left(\frac{dy}{dx}\right)-y=0$
• D. $y\left(\frac{dy}{dx}\right)+2x\left(\frac{dy}{dx}\right)-y=0$

Answer 1

Answer: (C)

$y^2=4ax+4a^2$
differentiate with respect to $x$
$\Rightarrow 2y\frac{dy}{dx}=4a$
$\Rightarrow a=\left(\frac{y}{2}\frac{dy}{dx}\right)$
so, required differential equation is
$y^2=\left(4\times \frac{y}{2}\frac{dy}{dx}\right)x+4{\left(\frac{y}{2}\frac{dy}{dx}\right)}^2$
$\Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2+2xy\left(\frac{dy}{dx}\right)-y^2=0$
$\Rightarrow y{\left(\frac{dy}{dx}\right)}^2+2x\left(\frac{dy}{dx}\right)-y=0$