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If the line $\mathrm{a} x+\mathrm{b} y+\mathrm{c}=0$ is a normal to the curve $x y=1$, then
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1734 Upvotes
Verified Answer
The correct answer is:
$a>0, b < 0$
$\begin{aligned}
& x y=1 \\
\therefore \quad & y=\frac{1}{x} \\
\therefore & y^{\prime}=\frac{-1}{x^2}
\end{aligned}$
$\therefore \quad$ 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 $\mathrm{b} < 0, \mathrm{a}>0$
& 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 $\mathrm{b} < 0, \mathrm{a}>0$
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