Problem:

The line \[ a^2x + ay + 1 = 0 \] is normal to the curve \[ xy = 1 \]. Find possible values of \( a \in \mathbb{R} \).

Step 1: Slope of Line

Rewrite: \[ y = -a x - \frac{1}{a} \] → slope = \( -a \)

Step 2: Curve Derivative

\[ xy = 1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \] Slope of normal = \( \frac{x}{y} \)

Match Slopes

\[ -a = \frac{x}{y} \Rightarrow x = -a y \]

Plug into Curve

\[ xy = 1 \Rightarrow (-a y)(y) = 1 \Rightarrow y^2 = -\frac{1}{a} \]

For real \( y \), we need \( a < 0 \)

✅ Final Answer:

\[ \boxed{a < 0} \]