The problem involves finding properties of tangents to a parametric curve.

Finding Parameter 't' for Point P

The curve is given by $x = 1 - 3t^2$ and $y = t - 3t^3$. We need to find the value of $t$ corresponding to the point P$(-2, 2)$.

• Set $x = -2$: $1 - 3t^2 = -2 \implies 3t^2 = 3 \implies t^2 = 1 \implies t = \pm 1$.
• Set $y = 2$: $t - 3t^3 = 2$.
• Test $t=1$: $1 - 3(1)^3 = 1 - 3 = -2 \ne 2$.
• Test $t=-1$: $(-1) - 3(-1)^3 = -1 - 3(-1) = -1 + 3 = 2$.

Thus, point P corresponds to $t = -1$. Let $t_P = -1$.

Calculating the Derivative dy/dx

First, find the derivatives of $x$ and $y$ with respect to $t$:

• $\frac{dx}{dt} = \frac{d}{dt}(1 - 3t^2) = -6t$
• $\frac{dy}{dt} = \frac{d}{dt}(t - 3t^3) = 1 - 9t^2$

The slope of the tangent, $\frac{dy}{dx}$, is given by:

$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 - 9t^2}{-6t} $

Slope of Tangent at P

Evaluate the slope at $t = t_P = -1$:

$ m_P = \frac{dy}{dx}\bigg|_{t=-1} = \frac{1 - 9(-1)^2}{-6(-1)} = \frac{1 - 9}{6} = \frac{-8}{6} = -\frac{4}{3} $

Tangent Line Equation and Intersection Point Q

The equation of the tangent line at P$(-2, 2)$ is $y - y_1 = m(x - x_1)$:

$ y - 2 = -\frac{4}{3}(x - (-2)) $

$ y - 2 = -\frac{4}{3}(x + 2) $

$ 3(y - 2) = -4(x + 2) $

$ 3y - 6 = -4x - 8 $

$ 4x + 3y + 2 = 0 $

To find where this tangent meets the curve again (point Q), substitute the parametric equations into the line equation:

$ 4(1 - 3t^2) + 3(t - 3t^3) + 2 = 0 $

$ 4 - 12t^2 + 3t - 9t^3 + 2 = 0 $

$ -9t^3 - 12t^2 + 3t + 6 = 0 $

Divide by $-3$:

$ 3t^3 + 4t^2 - t - 2 = 0 $

We know $t = -1$ is a root (point P). We can factor $(t+1)$ out:

$ (t+1)(3t^2 + t - 2) = 0 $

Factor the quadratic $3t^2 + t - 2 = 0$:

$ (3t - 2)(t + 1) = 0 $

So the roots are $t = -1$ (point P) and $t = 2/3$. The other intersection point Q corresponds to $t = 2/3$. Let $t_Q = 2/3$.

Slope of Tangent at Q

Evaluate the slope at $t = t_Q = 2/3$:

$ m_Q = \frac{dy}{dx}\bigg|_{t=2/3} = \frac{1 - 9(2/3)^2}{-6(2/3)} = \frac{1 - 9(4/9)}{-4} = \frac{1 - 4}{-4} = \frac{-3}{-4} = \frac{3}{4} $

Perpendicular Tangents Verification

Two tangents are perpendicular if the product of their slopes is $-1$.

$ m_P \times m_Q = \left(-\frac{4}{3}\right) \times \left(\frac{3}{4}\right) = -1 $

Since the product of the slopes is $-1$, the tangents at P and Q are at right angles.