Question

The differential equation(s) of family of curves whose tangent form an angle of $\frac{\pi }{4}$ with the hyperbola $xy=c^2$ is/are given by

• A. $\frac{dy}{dx}=\frac{x-y}{x+y}$
• B. $\frac{dy}{dx}=\frac{x}{x-y}$
• C. $\frac{dy}{dx}=\frac{x+y}{y-x}$
• D. $\frac{dy}{dx}=\frac{x}{y-x}$

Answer 1

Answer: (A), (C)

$\frac{dy}{dx}=\frac{x+y}{y-x}$

$xy=c^2$
$\Rightarrow x\frac{dy}{dx}+y=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}=m_2\text{(let)}$
Let $m_2$ be the slope of family of curves
$\therefore \tan \frac{\pi }{4}=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$1=\left|\frac{-\frac{y}{x}-m_2}{1-\frac{y}{x}{m}_2}\right|$
$\Rightarrow \frac{y}{x}+m_2=1-\frac{y}{x}{m}_2\text{or}\frac{y}{x}+m_2=\frac{y}{x}{m}_2-1$
$\Rightarrow m_2=\frac{1-\frac{y}{x}}{1+\frac{y}{x}}\text{or}{m}_2=\frac{\frac{y}{x}+1}{\frac{y}{x}-1}$
$\Rightarrow \frac{dy}{dx}=\frac{x-y}{x+y}\text{or}\frac{dy}{dx}=\frac{x+y}{y-x}$