If the line ax + by + 5 = 0 is a normal to th | vedclass

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(A) $1$. The curve is $y = \frac{1}{x}$. The slope of the tangent at any point $(x_1, y_1)$ is $\frac{dy}{dx} = -\frac{1}{x_1^2} = -y_1^2$.$2$. The sl

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$a > 0, b < 0$

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(A) $1$. The curve is $y = \frac{1}{x}$. The slope of the tangent at any point $(x_1, y_1)$ is $\frac{dy}{dx} = -\frac{1}{x_1^2} = -y_1^2$.$2$. The slope of the normal at $(x_1, y_1)$ is $m_n = -\frac{1}{dy/dx} = \frac{1}{y_1^2} = x_1^2$.$3$. The equation of the normal at $(x_1, y_1)$ is $y - y_1 = x_1^2(x - x_1)$, which simplifies to $x_1^2 x - y + (y_1 - x_1^3) = 0$.$4$. Comparing this with $ax + by + 5 = 0$, we have $\frac{x_1^2}{a} = \frac{-1}{b} = \frac{y_1 - x_1^3}{5}$.$5$. Since $x_1^2 > 0$ and $a = -b x_1^2$, for the normal to exist, $a$ and $b$ must have opposite signs. Specifically, since $x_1^2 > 0$, $a$ and $b$ must satisfy $a/b = -x_1^2  0$ and $b < 0$ is the correct condition.

If the line $ax + by + 5 = 0$ is a normal to the curve $xy = 1$, then which of the following is true?

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