<Question>
Find the curve defined by the equation
\[r^2 \cos 2 \theta = 4.\](A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola

Enter the letter of the correct option.
</Question>

<think>
<Plan_1>
Okay, so I need to figure out what kind of curve the equation \( r^2 \cos 2\theta = 4 \) represents. The options are line, circle, parabola, ellipse, or hyperbola. Hmm, let's start by recalling some polar coordinate equations and see if I can convert this into Cartesian coordinates to identify the conic section.

First, I remember that in polar coordinates, \( r \) is the radius, \( \theta \) is the angle, and the relationships between polar and Cartesian coordinates are \( x = r \cos \theta \), \( y = r \sin \theta \), and \( r^2 = x^2 + y^2 \). Also, double angle identities might come in handy here since there's a \( \cos 2\theta \) term.

The given equation is \( r^2 \cos 2\theta = 4 \). Let me try to manipulate this equation step by step. Maybe I can express \( \cos 2\theta \) using a double angle formula.
The double angle formula for cosine is \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). Alternatively, it can also be written as \( 2\cos^2 \theta - 1 \) or \( 1 - 2\sin^2 \theta \). Not sure which form will be more useful yet, but let's try substituting one of these into the equation.

So substituting \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \) into the equation gives:
\[ r^2 (\cos^2 \theta - \sin^2 \theta) = 4 \]

But \( r^2 \cos^2 \theta \) is \( (r \cos \theta)^2 = x^2 \), and similarly \( r^2 \sin^2 \theta = y^2 \). Therefore, the equation becomes:
\[ x^2 - y^2 = 4 \]

Wait, that's a hyperbola! The standard form of a hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

<prompt_1.1> Compare the derived equation with the standard form of a hyperbola and identify the correct option. </prompt_1.1>

But let me double-check to make sure I didn't make any mistakes in the substitution. Let's go through the steps again.

<prompt_1.2> Re-evaluate the steps to verify the conversion from polar to Cartesian coordinates. </prompt_1.2>

Alternatively, maybe I can approach this by converting the original equation into Cartesian coordinates without using the double angle identity. Let's see.

<prompt_1.3> Explore an alternative method to convert the original equation into Cartesian coordinates. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
In this case, \( a^2 = 4 \) and \( b^2 = 4 \), so \( a = 2 \) and \( b = 2 \). Therefore, the equation \( x^2 - y^2 = 4 \) is a hyperbola centered at the origin with vertices at (±2, 0).

Wait, but let me confirm if this conversion is correct. Let's start from the original equation:

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

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

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

Then, \( r^2 \cos^2 \theta = x^2 \), \( r^2 \sin^2 \theta = y^2 \), so substituting those in:

\( x^2 - y^2 = 4 \). Yes, that's correct.
</execution_1.1>

<execution_1.2>
Starting with \( r^2 \cos 2\theta = 4 \). Express \( \cos 2\theta \) as \( \cos^2 \theta - \sin^2 \theta \). Then:

Left-hand side (LHS) becomes \( r^2 (\cos^2 \theta - \sin^2 \theta) \). Then, since \( r \cos \theta = x \) and \( r \sin \theta = y \), we can write \( r^2 \cos^2 \theta = x^2 \) and \( r^2 \sin^2 \theta = y^2 \). Therefore, substituting these into the equation:

\( x^2 - y^2 = 4 \).
</execution_1.2>

<execution_1.3>
Starting with \( r^2 \cos 2\theta = 4 \). Let's express \( \cos 2\theta \) in terms of Cartesian coordinates. Since \( \cos \theta = x/r \) and \( \sin \theta = y/r \), then \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta = (x^2/r^2) - (y^2/r^2) = (x^2 - y^2)/r^2 \). Therefore, substituting back into the equation:

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

Simplifying the left side, the \( r^2 \) cancels out, leaving \( x^2 - y^2 = 4 \). Yep, that's the same result as before. So that's a hyperbola.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But let me think again—is there another way this could be represented? For example, if I had used a different form of the double angle identity. Let's try using \( \cos 2\theta = 2\cos^2 \theta - 1 \).

<prompt_2.1> Use an alternative form of the double angle identity to convert the original equation into Cartesian coordinates. </prompt_2.1>

Based on execution_1.2:
Alternatively, maybe I can test some points. Let's choose specific angles and see where the curve passes through.

<prompt_2.2> Test specific angles to verify the curve defined by the equation. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Then the equation becomes:

\( r^2 (2\cos^2 \theta - 1) = 4 \)

But then, how would that help? Let's try to express this in terms of x and y. Since \( \cos \theta = x/r \), so \( \cos^2 \theta = x^2/r^2 \). Therefore, substituting back:

\( r^2 \left(2 \frac{x^2}{r^2} - 1\right) = 4 \)

Simplify inside the brackets: \( 2x^2/r^2 - 1 \). Multiply by \( r^2 \):

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

But \( r^2 = x^2 + y^2 \), so substitute that in:

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

Simplify:

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

Same result. So regardless of the identity used, we arrive at the same Cartesian equation.
</execution_2.1>

<execution_2.2>
For example, when \( \theta = 0 \), \( \cos 0 = 1 \), so the equation becomes \( r^2 \times 1 = 4 \), so \( r^2 = 4 \), which gives \( r = 2 \). So in polar coordinates, that's (2, 0), which converts to (2, 0) in Cartesian.

When \( \theta = \pi/4 \), \( \cos(2 * \pi/4) = \cos(\pi/2) = 0 \), so the equation becomes \( r^2 * 0 = 4 \), which implies 0 = 4. That's impossible, so there's no point at \( \theta = \pi/4 \). Makes sense because in Cartesian, when \( \theta = \pi/4 \), that's the line y = x, and the hyperbola \( x^2 - y^2 = 4 \) does intersect the x-axis at (±2, 0) but doesn't cross y = x because plugging y = x into the hyperbola equation gives \( x^2 - x^2 = 4 \) which is 0 = 4, impossible. So that's consistent.

Another angle: \( \theta = \pi/3 \). Then \( 2\theta = 2\pi/3 \), and \( \cos(2\pi/3) = -1/2 \). So the equation becomes \( r^2 (-1/2) = 4 \), so \( r^2 = -8 \). But \( r^2 \) can't be negative, so no solution. So in polar coordinates, there's no point at \( \theta = \pi/3 \). That also makes sense with the hyperbola, since the hyperbola \( x^2 - y^2 = 4 \) only exists in regions where x is large enough. Specifically, for real points, \( x^2 \geq 4 \), so |x| ≥ 2.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Comparing this to the general form of a conic section in polar coordinates, which is \( r = \frac{e d}{1 + e \cos \theta} \) or something similar, but this equation is different.
Alternatively, in polar coordinates, conic sections are usually expressed with \( r \) in the numerator, but here we have \( r^2 \) multiplied by a cosine term.

But regardless of the form, converting to Cartesian coordinates seems to be the way to go here.
Since we arrived at \( x^2 - y^2 = 4 \), which is a hyperbola, that should be the answer.

But let me check if any other form could represent a different conic.

<prompt_3.1> Consider alternative forms and confirm the type of conic section. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
For example, if it were a circle, the equation would be \( x^2 + y^2 = r^2 \), but here we have \( x^2 - y^2 = 4 \), which is a hyperbola. Parabola would have a different form, like \( y^2 = 4ax \) or something. Ellipse is \( x^2/a^2 + y^2/b^2 = 1 \). So this equation doesn't fit any of those.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, the answer should be (E) Hyperbola. Let me just cross-verify once again.

<prompt_4.1> Perform a final verification to confirm the solution. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Original equation: \( r^2 \cos 2\theta = 4 \). Convert to Cartesian:

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

Express \( \cos 2\theta \) as \( (x^2 - y^2)/r^2 \):

\( r^2 \times (x^2 - y^2)/r^2 = 4 \)

Simplifies to \( x^2 - y^2 = 4 \). Yep, that's a hyperbola. So (E) is the correct answer.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
**Final Answer**
\boxed{E}
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the curve defined by the equation \(r^2 \cos 2 \theta = 4\) is a hyperbola, and the correct answer is \(\boxed{E}\).
</Final_answer>
</think>