Question

The differential equation of the family of curves
$\frac{x^2}{a^2}+\frac{y^2}{a^2+{\lambda }^2}=1$ is ($\lambda$ is orbitary constant)

• A. $(x^2-a^2)y_1=xy$
• B. $(x^2-a^2)y_2-xy=0$
• C. $x^2{y}_2-a^{2}y=0$
• D. $(x^2-a^2)y_1+xy=0$

Answer 1

The correct option is A $(x^2-a^2)y_1=xy$

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

$\Rightarrow \frac{2x}{a^2}+\frac{2y{y}_1}{a^2+{\lambda }^2}=0$

$\Rightarrow -\frac{x}{a^2}=\frac{y{y}_1}{a^2+{\lambda }^2}$

$\Rightarrow a^2+{\lambda }^2=-\frac{a^{2}y{y}_1}{x}$

$\Rightarrow {\lambda }^2=-a^2(1+\frac{y{y}_1}{x})$

From $\left(1\right)$,

$\Rightarrow \frac{x^2}{a^2}-\frac{y^{2}x}{a^{2}y{y}_1}=1$

$\Rightarrow y{y}_1{x}^2-y^{2}x=a^{2}y{y}_1$

$\Rightarrow (x^2-a^2)y_1=xy$