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:

$$\boxed{a < 0}$$

Finding Foci of a Conic

Given Equation: $x^2 + 2x - 4y^2 + 8y - 7 = 0$

Step 1: Complete the square

⇒ $(x + 1)^2 - 4(y - 1)^2 = 4$

Rewriting: $\frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1$

This is a horizontal hyperbola with:

• Center: $(-1, 1)$
• $a^2 = 4$, $b^2 = 1$
• $c = \sqrt{a^2 + b^2} = \sqrt{5}$

✅ Foci: $(-1 \pm \sqrt{5},\ 1)$