Question

A family of curves is given by the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The differential equation representing this family of curves is given by $x y \frac{d^{2} y}{d x^{2}}+A x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$. The value of $A$ is

• A. 0
• B. 1
• C. 3
• D. 5

Answer 1

Answer: (B)

Differentiating the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
w.r.t. $x$, we get
$\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x}=0$ or
$\frac{x^{2}}{a^{2}}+\frac{x y}{b^{2}} \frac{d y}{d x}=0$
or $1-\frac{y^{2}}{b^{2}}+\frac{x y}{b^{2}} \frac{d y}{d x}=0$
$\left[\because \frac{x^{2}}{a^{2}}=1-\frac{y^{2}}{b^{2}}\right]$
Differentiating again w.r.t. $x$, we get
$\frac{-2 y}{b^{2}} \frac{d y}{d x}+\frac{y}{b^{2}} \frac{d y}{d x}+\frac{x}{b^{2}} \frac{d y}{d x} \cdot \frac{d y}{d x}+\frac{x y}{b^{2}} \frac{d^{2} y}{d x^{2}}=0$
or $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$
comparing with the given differential equation, we get $A =1$.