Let's trace the curve given by the polar equation: \[ r = \frac{2a}{1 + \cos \theta} \] ### **Step 1: Identify the Type of Curve** This equation is of the form: \[ r = \frac{l}{1 + e \cos \theta} \] where \( l = 2a \) and \( e = 1 \). This is the standard polar form of a **conic with the focus at the pole**. For \( e = 1 \), the curve is a **parabola**. --- ### **Step 2: Symmetry** - The equation involves only \( \cos\theta \), which is an even function. - Therefore, the curve is **symmetric about the initial line** (\(\theta = 0\)). --- ### **Step 3: Special Points** #### (a) For \( \theta = 0 \): \[ r = \frac{2a}{1 + 1} = \frac{2a}{2} = a \] So, the curve passes through the point \( (a, 0) \). #### (b) For \( \theta = \pi \): \[ r = \frac{2a}{1 + (-1)} = \frac{2a}{0} \] So, \( r \to \infty \) as \( \theta \to \pi \). #### (c) For \( \theta = \frac{\pi}{2} \): \[ r = \frac{2a}{1 + 0} = 2a \] So, the curve passes through \( (2a, \frac{\pi}{2}) \). #### (d) For \( \theta = -\frac{\pi}{2} \): \[ r = \frac{2a}{1 + 0} = 2a \] So, the curve passes through \( (2a, -\frac{\pi}{2}) \). --- ### **Step 4: Cartesian Form (Optional)** Let's convert to Cartesian coordinates: Recall: \[ r = \sqrt{x^2 + y^2}, \quad \cos\theta = \frac{x}{r} \] So, \[ r = \frac{2a}{1 + \cos\theta} \implies r(1 + \cos\theta) = 2a \] \[ r + r\cos\theta = 2a \] \[ r + x = 2a \quad (\text{since } r\cos\theta = x) \] \[ \sqrt{x^2 + y^2} + x = 2a \] \[ \sqrt{x^2 + y^2} = 2a - x \] \[ x^2 + y^2 = (2a - x)^2 \] \[ x^2 + y^2 = 4a^2 - 4ax + x^2 \] \[ y^2 = 4a^2 - 4ax \] or \[ y^2 = 4a(a - x) \] This is the equation of a **parabola** that opens to the left. --- ### **Step 5: Sketch & Properties** - **Vertex:** At \( (a, 0) \) - **Axis:** Along the x-axis (leftwards) - **Focus:** At the pole (origin) - **Directrix:** \( x = 2a \) - As \( \theta \to \pi \), \( r \to \infty \), so the parabola extends infinitely to the left. --- ## **Final Answer** ### The curve \( r = \frac{2a}{1 + \cos\theta} \) is a parabola with: - Focus at the pole (origin), - Vertex at \( (a, 0) \), - Axis along the x-axis (leftwards), - Directrix at \( x = 2a \), - Cartesian equation: \(\boxed{y^2 = 4a(a - x)}\). **Rough sketch:** ``` ^ | | * (vertex at (a, 0)) | * | * ------|----------------------> | * | * | * ```

Let's carefully analyze the given polar equation step-by-step, understand its geometric nature, and derive its Cartesian form. The goal is to trace the curve described by: \[ r = \frac{2a}{1 + \cos \theta} \] --- ### **Step 1: Recognize the type of curve** The given polar equation is in the form: \[ r = \frac{l}{1 + e \cos \theta} \] where: - \( l = 2a \) - \( e = 1 \) This is the standard form of a conic section with focus at the pole (origin). - When \( e < 1 \), it's an ellipse. - When \( e = 1 \), it's a parabola. - When \( e > 1 \), it's a hyperbola. Since \( e = 1 \), **the curve is a parabola** with focus at the pole. --- ### **Step 2: Find key points on the curve** Let's compute some specific points for particular angles \( \theta \): --- **(a) At \( \theta = 0 \):** \[ r = \frac{2a}{1 + \cos 0} = \frac{2a}{1 + 1} = \frac{2a}{2} = a \] *Coordinates:* \[ x = r \cos \theta = a \times 1 = a \] \[ y = r \sin \theta = a \times 0 = 0 \] **Point:** \( (a, 0) \). --- **(b) At \( \theta = \pi \):** \[ r = \frac{2a}{1 + \cos \pi} = \frac{2a}{1 - 1} = \frac{2a}{0} \to \infty \] *Interpretation:* As \( \theta \to \pi \), \( r \to \infty \). The curve extends infinitely in the direction \( \theta = \pi \) (to the left). --- **(c) At \( \theta = \frac{\pi}{2} \):** \[ r = \frac{2a}{1 + 0} = 2a \] *Coordinates:* \[ x = 2a \times \cos \frac{\pi}{2} = 0 \] \[ y = 2a \times \sin \frac{\pi}{2} = 2a \] Point: \( (0, 2a) \). --- **(d) At \( \theta = - \frac{\pi}{2} \):** \[ r = \frac{2a}{1 + 0} = 2a \] \[ x = 2a \times \cos \left(- \frac{\pi}{2}\right) = 0 \] \[ y = 2a \times \sin \left(- \frac{\pi}{2}\right) = - 2a \] Point: \( (0, -2a) \). --- ### **Step 3: Convert the polar equation to Cartesian form** Recall: \[ r = \sqrt{x^2 + y^2}, \quad \cos \theta = \frac{x}{r} \] Substitute into the original: \[ r = \frac{2a}{1 + \frac{x}{r}} \] Multiply both sides by denominator: \[ r (1 + \frac{x}{r}) = 2a \] Simplify: \[ r + x = 2a \] Replace \( r = \sqrt{x^2 + y^2} \): \[ \sqrt{x^2 + y^2} + x = 2a \] **Isolate the square root:** \[ \sqrt{x^2 + y^2} = 2a - x \] Since \( r \geq 0 \), the right side must be non-negative: \[ 2a - x \geq 0 \implies x \leq 2a \] Now, square both sides: \[ x^2 + y^2 = (2a - x)^2 \] Expand RHS: \[ x^2 + y^2 = 4a^2 - 4a x + x^2 \] Subtract \( x^2 \) from both sides: \[ y^2 = 4a^2 - 4a x \] **Final Cartesian form:** \[ \boxed{ y^2 = 4a (a - x) } \] --- ### **Step 4: Analyze the Cartesian equation** The equation: \[ y^2 = 4a(a - x) \] is a **parabola opening towards negative \( x \)** (since as \( x \to -\infty \), \( y^2 \to \infty \)). - **Vertex:** occurs when \( y = 0 \), at \( x = a \). At \( y=0 \): \[ 0 = 4a(a - x) \implies x = a \] So, **vertex at \( (a, 0) \)**. --- ### **Step 5: Geometrical properties** - The focus is at the pole \( O(0, 0) \). - The parabola has **vertex at \( (a, 0) \)**. - The **directrix** is a vertical line, which can be deduced from the standard form for parabola: From the Cartesian form: \[ y^2 = 4a (a - x) \] which matches the standard parabola form: \[ (y - 0)^2 = 4 p (x - h) \] with: \[ h = a, \quad 4 p = -4a \implies p = -a \] Since \( p < 0 \), the parabola opens **towards decreasing \( x \)** (to the left). The focus is located at: \[ \text{Focus} = (h + p, 0) = (a - a, 0) = (0, 0) \] which is at the pole (origin), confirming the focus is at the pole. The **directrix** is at: \[ x = h - p = a - (-a) = a + a = 2a \] --- ### **### Final summarized answer:** **The given polar curve:** \[ \boxed{ r = \frac{2a}{1 + \cos \theta} } \] **is a parabola with:** - **Focus at the pole \( (0, 0) \).** - **Vertex at \( (a, 0) \).** - **Axis along the x-axis, opening leftward.** - **Directrix at \( x = 2a \).** - **Cartesian equation:** \[ \boxed{ y^2 = 4a (a - x) } \] This completes the detailed step-by-step analysis of the curve.