Question

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

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

Answer 1

The correct option is A $(x-\frac{y}{y^{\prime }})(x+y{y}^{\prime })=a^2-b^2$

$\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1$
Differentiating we get,
$\frac{2x}{a^2+\lambda }+\frac{2y}{b^2\lambda }{y}^{\prime }=0$
or $x(b^2+\lambda )+y{y}^{\prime }\left(a^2\lambda \right)=0$
$\therefore \lambda =-\frac{b^{2}x+a^{2}y{y}^{\prime }}{x+y{y}^{\prime }}$
Hence $a^2+\lambda =a^2-\frac{b^{2}x+a^{2}y{y}^{\prime }}{x+y{y}^{\prime }}=\frac{(a^2-b^2)}{x+y{y}^{\prime }}$
and $b^2+\lambda =b^2-\frac{b^{2}x+a^{2}y{y}^{\prime }}{x+y{y}^{\prime }}=-\frac{(a^2-b^2)}{x+y{y}^{\prime }}$
Putting the values of $a^2+\lambda$
and $b^2+\lambda$ in the given equation.
$(x-\frac{y}{y^{\prime }})(x+y{y}^{\prime })=a^2-b^2$