Question

Obtain the differential equation of the family of parabola having their focus at the origin and the axis along the x-axis.

Answer 1

Let direction be $x=a$

so, vertex will be $(\frac{a}{2},0)$

Then, equation will be

$a^2-2ax-y^2=0$

differentiating it, we get

$-2a-2y\frac{dy}{dx}=0a=-y\frac{dy}{dx}$

Put $a$ in original equation we get

${(-y\frac{dy}{dx})}^2-2(-y\frac{dy}{dx})\left(x\right)-y^2=0y^2{\left(\frac{dy}{dx}\right)}^2+2xy\frac{dy}{dx}+x^2=x^2+y^2{\left(y\frac{dy}{dx}\right)}^2+2x\left(y\frac{dy}{dx}\right)+x^2=x^2+y^2{(y\frac{dy}{dx}+x)}^2=x^2+y^2$

This is the required equation.