If the line ax + by + c =0 is a normal to the curve xy = 1, then

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 or a < 0, b > 0 (C) a < 0, b < 0 (D) none of these.

Tardigrade Admin · Accepted Answer

Since m of the normal is always +ve in this case a > 0, b < 0 or a < 0, b > 0 ( textslope m= -(a/b)>0) [y=(1/x) ∴ (dy/dx)= -(1/x2) ∴ -(1/dy/dx ) = x2>0 for all x ∴ slope of normal is > 0]

Q. If the line $a x + b y + c = 0$ is a normal to the curve $x y = 1$ , then

$a > 0, b > 0$

$a > 0, b < 0$ or $a < 0, b > 0$

$a < 0, b < 0$

none of these.

Since $m$ of the normal is always $+ v e$ in this case
$a > 0, b < 0$ or $a < 0, b > 0$
$(slope m = - \frac{a}{b} > 0)$
[ $y = \frac{1}{x} ∴ \frac{d y}{d x} = - \frac{1}{x ^{2}} ∴ - \frac{1}{d y / d x} = x^{2} > 0$
for all $x ∴$ slope of normal is $> 0$ ]

Q. If the line ax+by+c=0 is a normal to the curve xy=1, then

a>0,b>0

a>0,b<0 or a<0,b>0

a<0,b<0