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. a0
• D. a < 0, b < 0

Answer 1

The correct option is C a0

Differentiating the equation of curve xy = 1
We have x$\frac{dy}{dx}$ + y = 0 $\Rightarrow \frac{dy}{dx}$ = -$\frac{y}{x}$
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 $<$t 0, b $>$ 0.