Question

Consider the equation $\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1$ where a and b are specified constants and $\lambda$ is an arbitrary parameter. Find a differential equation satisfied by it.

Answer 1

$\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1$........(i)

$\Rightarrow \frac{2x}{a^2+\lambda }+\frac{2y}{b^2+\lambda }\frac{dy}{dx}=0$......(ii)

$\Rightarrow \frac{2}{a^2+\lambda }+\frac{2}{b^2+\lambda }{\left(\frac{dy}{dx}\right)}^2+\frac{2y}{b^2+\lambda }.\frac{d^{2}y}{d{x}^2}=0$

$\Rightarrow \frac{2}{a^2+\lambda }=-\frac{2}{b^2+\lambda }{\left(\frac{dy}{dx}\right)}^2-\frac{2y}{b^2+\lambda }\frac{d^{2}y}{d{x}^2}$......(iii)

Substituting eq. (iii) in eq. (ii) , we get

$\Rightarrow [-\frac{2}{b^2+\lambda }{\left(\frac{dy}{dx}\right)}^2-\frac{2y}{b^2+\lambda }\frac{d^{2}y}{d{x}^2}]x+\frac{2y}{b^2+\lambda }\frac{dy}{dx}=0$

$\Rightarrow y.\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=y.\frac{dy}{dx}$

$\Rightarrow y{y}_2+(y_1{)}^2-y{y}_1=0$