Step 1: Understanding the Concept:
We analyze the symmetry of a parametric curve by replacing \( t \) with \( -t \). For the tangent geometry, we calculate the slope at point P, find where it intersects the curve again at Q, and then compare the slopes at P and Q.
Step 2: Key Formula or Approach:
1. Symmetry: Replace \( t \to -t \).
2. Slope \( m = \frac{dy/dt}{dx/dt} = \tan \phi \).
Step 3: Detailed Explanation:
Symmetry:
Replace \( t \) with \( -t \): \( x_{new} = 1 - 3(-t)^2 = 1 - 3t^2 = x \).
\( y_{new} = (-t) - 3(-t)^3 = -t + 3t^3 = -y \).
Since \( (x, y) \to (x, -y) \), the curve is symmetric about the x-axis. (A) is correct.
Tangent at P:
Point \( P(-2, 2) \). For \( x = -2 \implies 1 - 3t^2 = -2 \implies 3t^2 = 3 \implies t = \pm 1 \).
For \( y = 2 \implies t - 3t^3 = 2 \implies t(1 - 3t^2) = 2 \).
Using \( t=1 \): \( 1(1-3) = -2 \). No.
Using \( t=-1 \): \( -1(1-3) = 2 \). Yes. So point P corresponds to \( t = -1 \).
Slope \( m = \frac{1 - 9t^2}{-6t} \). At \( t = -1 \), \( m_P = \frac{1 - 9}{6} = -\frac{8}{6} = -\frac{4}{3} \).
Meeting the curve again at Q:
The tangent line at P is \( y - 2 = -\frac{4}{3}(x + 2) \implies 3y + 4x + 2 = 0 \).
Substitute parametric forms: \( 3(t - 3t^3) + 4(1 - 3t^2) + 2 = 0 \).
\[ 3t - 9t^3 + 4 - 12t^2 + 2 = 0 \implies 9t^3 + 12t^2 - 3t - 6 = 0 \]
Since \( t = -1 \) is the tangency point, \( (t+1)^2 \) is a factor.
Factoring: \( (t+1)^2 (9t - 6) = 0 \implies t = 2/3 \) for point Q.
Slope at Q (\( t = 2/3 \)): \( m_Q = \frac{1 - 9(4/9)}{-6(2/3)} = \frac{1 - 4}{-4} = \frac{3}{4} \).
Product of slopes \( m_P \cdot m_Q = (-4/3)(3/4) = -1 \).
Tangents are at right angle. (D) is correct.
Step 4: Final Answer:
Options (A) and (D) are correct.