Given the equation \( a^2 x + a y + 1 = 0 \), we are asked to determine the condition for when the line is normal to the curve \( x y = 1 \). \[\] Start with the equation of the curve \( x y = 1 \). Implicit differentiation with respect to \( x \): \[ x \frac{dy}{dx} + y = 0 \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-y}{x} \] The slope of the normal line is the negative reciprocal of the slope of the tangent. Hence, the slope of the normal is: \[ \text{Slope of normal} = \frac{x}{y} \] substitute the given equation \( y = \frac{1}{x} \) \[ \frac{x}{y} = x^2 \] From this, we have: \( y = -ax - \frac{1}{a} \) \[ x^2 = -a \] For this equation to hold for real values of \( x \), we must have \( a < 0 \).