Q93 / 150 · MHT CET 2023 (10 May Shift 1) · Mathematics
If the line $\mathrm{a} x+\mathrm{b} y+\mathrm{c}=0$ is a normal to the curve $x y=1$, then
- A$a>0, b>0$
- Correct answerB$a>0, b < 0$
- C$\mathrm{a} < 0, \mathrm{~b} < 0$
- D$a=0, b=0$
Solution
$\begin{aligned}
& x y=1 \\
\therefore \quad & y=\frac{1}{x} \\
\therefore & y^{\prime}=\frac{-1}{x^2}
\end{aligned}$
$\therefore \quad$ Slope of the normal $=x^2$
Slope of the line $a x+b y+c=0$ is $\frac{-a}{b}$.
Since the line $a x+b y+c=0$ is a normal to the curve $x y=1$,
$x^2=-\frac{\mathrm{a}}{\mathrm{b}}$
For this condition to hold true, either $\mathrm{a} < 0, \mathrm{~b}>0$ or $\mathrm{b} < 0, \mathrm{a}>0$
Hence, the correct answer is option: 'B'.