Question

If the line $aX + bY + c = 0$ is a normal to the curve xy =1. Then

• A. a > 0, b >0
• B. a >0, b < 0
• C. a <0, b>0
• D. a < 0, b < 0

Answer 1

Answer: (C)

a <0, b>0

Answer 2

Differentiating the equation of curve $xy = 1$

We have $x\frac{dy}{dx} + y = 0$ ⇒ $\frac{dy}{dx} = - \frac{y}{x}$

Hence the slope of normal = $\frac{x}{y}$.

Moreover the slope of the line $aX + bY + c = 0$ is $- \frac{a}{b}$. So we have $\frac{x}{y} = - \frac{a}{b}$, i.e. $bx + ay = 0$ solving this with $xy = 1$, we have $x^{2} = - \frac{a}{b}$ So we must have $\frac{a}{b} < 0$

i.e., $a > 0,b < 0$ or $a < 0,b > 0$.