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