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Question
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:
\[ \boxed{a < 0} \]
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