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

Answer: (A), (B)

$a>0,b<0$

Explanation for the correct options:

The given straight line, $L_1:ax+by+c=0$.

The given equation of the curve, $C_1:xy=1$.

Differentiate the equation of $C_1$ with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow y\frac{dx}{dx}+x\frac{dy}{dx}=0 \\ \Rightarrow y+x\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{y}{x} \\ \Rightarrow \frac{dy}{dx}=-\frac{xy}{x^2} \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{x^2}\left[\because xy=1\right] \end{gathered}$$

So, the slope of the curve at the point $\left(h,k\right)$ can be given by $m_1=-\frac{1}{h^2}$.

Therefore, the slope of the normal of $C_1$ can be given by, $-\frac{1}{m_1}=h^2$.

The equation of $L_1$ can be represented as, $y=-\frac{a}{b}x-\frac{c}{b}$.

Compare the above equation with the slope-intercept form of a straight line, $y=mx+c$, where $m$ is slope.

Thus, the slope of the line $L_1$ is $m_2=-\frac{a}{b}$.

As $L_1$ is normal to the curve $C_1$, thus $m_2=h^2$

$-\frac{a}{b}=h^2$.

As $h^2>0$, so $\left(-\frac{a}{b}\right)>0$.

$\Rightarrow \frac{a}{b}<0$

Thus, $a$ and $b$ must be opposite in sign.

So, either $a<0,b>0$ or $a>0,b<0$.

Hence, (A) and (B) are correct options.