Question

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

• A. $a>0,b<0$
• B. $a>0,b>0$
• C. $a<0,b<0$
• D. Data is insufficient

Answer 1

The correct option is A

$a>0,b<0$

Explanation for the correct options:

Normal on the curve:

Given curve is $xy=1$

So, $y=\frac{1}{x}...\left(1\right)$

Differentiating with respect to $x$, we get,
$\begin{array}{ccc}& x\frac{\mathrm{d}y}{\mathrm{d}x}+y=0& \left[\because \left(uv\right)\text{'}=uv\text{'}+u\text{'}v\right]\\ \Rightarrow & \frac{dy}{dx}=-\frac{y}{x}& \end{array}$

Given line is, $ax+by+c=0$
$\Rightarrow y=-\frac{a}{b}x-\frac{c}{b}$

The slope of a line is the coefficient of $x$ in $y=mx+c$ form.
Thus, the slope of the given line is, $-\frac{a}{b}$
Thus the slope of tangent is, $\frac{b}{a},\mathbf{}\left[\because \text{slope of normal}=-\frac{1}{\text{slope of line}}\right]$

Since the slope is normal to the curve, the normal of the line will be tangent to the curve. Thus,

$\begin{array}{ccc}& -\frac{y}{x}=\frac{b}{a}& \\ \Rightarrow & -\frac{{\displaystyle \frac{1}{x}}}{x}=\frac{b}{a}& \left[\because y=\frac{1}{x}\right]\\ \Rightarrow & x^2=-\frac{a}{b}& \end{array}$

So, $\text{LHS}$ will always be a positive value, so $\frac{a}{b}$ in the $\text{RHS}$ must be negative. For that to happen, $a$ and $b$ must have the opposite signs.
Thus, $a>0,b<0$ or $a<0,b>0$.

Therefore, option A is correct.