Question

Form the differential equation of the family of curves represented by the equation : ${(x+a)}^2-2y^2=a^2.$

Answer 1

We have,

${(x-a)}^2+2y^2=a^2......\left(1\right)$

On differentiation with respect to $x$ and we get,

$\frac{d}{dx}[{(x-a)}^2+2y^2]=\frac{d}{dx}{a}^2$

$2(x-a)+4y\frac{dy}{dx}=0$

$(x-a)+2y\frac{dy}{dx}=0$

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

$a=2y\frac{dy}{dx}+x$

Put the value of $a$ in equation (1) and we get,

${[x-(2y\frac{dy/}{dx}+x)]}^2+2y^2={(2y\frac{dy}{dx}+x)}^2$

${[x-2y\frac{dy}{dx}-x]}^2+2y^2={(2y\frac{dy}{dx}+x)}^2$

${(-2y\frac{dy}{dx})}^2+2y^2=4y^2{\left(\frac{dy}{dx}\right)}^2+x^2+4xy\frac{dy}{dx}$

$2y^2=x^2+4xy\frac{dy}{dx}$

$2y^2-x^2-4xy\frac{dy}{dx}=0$

$2y^2-x^2=4xy\frac{dy}{dx}$

$4xy\frac{dy}{dx}=2y^2-x^2$

$\frac{dy}{dx}=\frac{2y^2-x^2}{4xy}$

Hence, this is the answer.