Question

Find the differential equation of the family of curves whose equations are $\frac{x^2}{a^2}+\frac{y^2}{a^2+\lambda }=1,$ where $\lambda$ is parameter.

• A. $-\frac{xy}{a^2{y}^{\prime }}=\frac{a^2-x^2}{a^2}$
• B. $-\frac{xy}{a^2{y}^{\prime }}=\frac{a^2+x^2}{a^2}$
• C. $-\frac{xy}{a^2{y}^{\prime }}=\frac{a^4-x^2}{a^2}$
• D. $-\frac{xy}{a^2{y}^{\prime }}=\frac{a^4+x^2}{a^2}$

Answer 1

The correct option is A $-\frac{xy}{a^2{y}^{\prime }}=\frac{a^2-x^2}{a^2}$

Differentiating given equation w.r.t $x$ we get
$\frac{2x}{a^2}+\frac{2y}{a^2+\lambda }\frac{dy}{dx}=0$
$\therefore (a^2+\lambda )=-\frac{a^{2}y{y}^{\prime }}{x}$
Putting for $a^2+\lambda$ in the given equation , we get
$\Rightarrow \frac{x^2}{a^2}-\frac{y^{2}x}{a^{2}y{y}^{\prime }}=1$

$\Rightarrow \frac{x^2}{a^2}-\frac{y^{2}x}{a^{2}y{y}^{\prime }}=1$
$\Rightarrow -\frac{xy}{a^2{y}^{\prime }}=1-\frac{x^2}{a^2}=\frac{a^2-x^2}{a^2}$