Alright, let me try to figure out this problem. The question is asking what kind of curve is defined by the equation \( r^2 \cos 2\theta = 4 \). The options are Line, Circle, Parabola, Ellipse, or Hyperbola. Hmm, okay. So, first, I need to recall what different conic sections look like in polar coordinates. 

I remember that in polar coordinates, the general form of a conic section is \( r = \frac{ed}{1 + e \cos \theta} \), where e is the eccentricity, d is the distance from the directrix, and θ is the angle. But this equation here is a bit different—it's \( r^2 \cos 2\theta = 4 \). So, maybe I need to convert this equation into Cartesian coordinates to see which conic section it represents. 

Let me recall the conversion formulas from polar to Cartesian coordinates. We have \( x = r \cos \theta \) and \( y = r \sin \theta \), so \( r^2 = x^2 + y^2 \). Also, \( \cos 2\theta \) can be expressed using the double-angle identity: \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). Let me write that down.

So, substituting \( \cos 2\theta \) into the original equation:

\( r^2 \cos 2\theta = 4 \)

Which becomes:

\( r^2 (\cos^2 \theta - \sin^2 \theta) = 4 \)

Now, substitute \( r^2 = x^2 + y^2 \), \( \cos \theta = x/r \), and \( \sin \theta = y/r \):

\( (x^2 + y^2) \left( \left( \frac{x}{r} \right)^2 - \left( \frac{y}{r} \right)^2 \right) = 4 \)

Simplify the terms inside the parentheses:

\( \left( \frac{x^2 - y^2}{r^2} \right) \)

So, substituting back:

\( (x^2 + y^2) \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = 4 \)

Oh, look at that! The \( x^2 + y^2 \) cancels out with the denominator, so we're left with:

\( x^2 - y^2 = 4 \)

Whoa, that's a hyperbola! Because the standard form of a hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), so this equation is similar but scaled by 4. So, if I divide both sides by 4:

\( \frac{x^2}{4} - \frac{y^2}{4} = 1 \)

Which is clearly a hyperbola with a = 2 and b = 2. Therefore, the curve is a hyperbola. So, the answer should be (E) Hyperbola.

Wait, but let me double-check my steps to make sure I didn't make a mistake. Starting from \( r^2 \cos 2\theta = 4 \), converted to Cartesian coordinates. Let me verify each substitution:

1. \( r^2 = x^2 + y^2 \) – correct.
2. \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \) – yes, that's the double angle formula.
3. Then substituting \( \cos \theta = x/r \) and \( \sin \theta = y/r \) – that's correct.
4. Then \( (x^2 + y^2) \times (\frac{x^2 - y^2}{x^2 + y^2}) = 4 \) – simplifies to \( x^2 - y^2 = 4 \) – right.

Yes, that seems right. So, the equation reduces to a hyperbola. Therefore, the answer is E. 

Alternatively, maybe there's another way to approach this. For example, recognizing that \( r^2 \cos 2\theta \) is part of the general conic section equation in polar form. Let me recall that in polar coordinates, the equation can also be written as \( r = \frac{ed}{1 + e \cos \theta} \) for an ellipse or hyperbola, but here it's \( r^2 \cos 2\theta = 4 \), which seems different. Wait, maybe using the identity \( \cos 2\theta = 2 \cos^2 \theta - 1 \) or \( 1 - 2 \sin^2 \theta \), but I already used the other identity. Either way, substituting and simplifying leads to the same result. 

Alternatively, another approach: express the original equation in terms of x and y. Let's try that again. 

Original equation: \( r^2 \cos 2\theta = 4 \)

Express \( \cos 2\theta \) as \( \cos^2 \theta - \sin^2 \theta \), so:

\( r^2 (\cos^2 \theta - \sin^2 \theta) = 4 \)

Now, express \( \cos^2 \theta \) as \( x^2 / r^2 \) and \( \sin^2 \theta \) as \( y^2 / r^2 \):

\( r^2 \left( \frac{x^2}{r^2} - \frac{y^2}{r^2} \right) = 4 \)

Simplify inside the parentheses:

\( \frac{x^2 - y^2}{r^2} \)

Multiply by \( r^2 \):

\( x^2 - y^2 = 4 \)

Same result. So, definitely a hyperbola. 

Therefore, the answer is (E) Hyperbola. 

I think that's thorough. I converted the equation step by step, checked the substitutions, and arrived at the conclusion. No mistakes detected in the process. The answer is E.
Therefore, the curve defined by the equation is a hyperbola.

\boxed{E}
</think>