If the tangent at a point P, with parameter $$t$$, on the curve $$x = 4t^2 + 3$$, $$y = 8t^3 - 1$$, $$t \in R$$, meets the curve again at a point Q, then the coordinates of Q are:

The curve is given parametrically by $$ x = 4t^2 + 3 $$ and $$ y = 8t^3 - 1 $$, where $$ t $$ is a real number. The point P on the curve corresponds to a specific parameter value, say $$ t $$, so its coordinates are $$ (4t^2 + 3, 8t^3 - 1) $$. The tangent at P intersects the curve again at another point Q, and we need to find the coordinates of Q.

To find the equation of the tangent at P, we first compute the derivative $$ \frac{dy}{dx} $$. Since the curve is parametric, we use $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$. Differentiating $$ x $$ and $$ y $$ with respect to $$ t $$:

$$ \frac{dx}{dt} = \frac{d}{dt}(4t^2 + 3) = 8t $$

$$ \frac{dy}{dt} = \frac{d}{dt}(8t^3 - 1) = 24t^2 $$

Thus,

$$ \frac{dy}{dx} = \frac{24t^2}{8t} = 3t \quad \text{for} \quad t \neq 0 $$

The slope of the tangent at P is $$ 3t $$. The equation of the tangent line at point $$ P(4t^2 + 3, 8t^3 - 1) $$ with slope $$ 3t $$ is given by:

$$ y - (8t^3 - 1) = 3t \left( x - (4t^2 + 3) \right) $$

Simplifying this equation:

$$ y - 8t^3 + 1 = 3t(x - 4t^2 - 3) $$

$$ y - 8t^3 + 1 = 3t x - 12t^3 - 9t $$

$$ y = 3t x - 12t^3 - 9t + 8t^3 - 1 $$

$$ y = 3t x - 4t^3 - 9t - 1 $$

This tangent line intersects the curve again at point Q. The curve is defined parametrically, so let the parameter at Q be $$ s $$. Thus, any point on the curve has coordinates $$ (4s^2 + 3, 8s^3 - 1) $$. Substituting these coordinates into the tangent line equation:

$$ 8s^3 - 1 = 3t(4s^2 + 3) - 4t^3 - 9t - 1 $$

Simplifying the right side:

$$ 3t(4s^2 + 3) = 12t s^2 + 9t $$

So,

$$ 8s^3 - 1 = 12t s^2 + 9t - 4t^3 - 9t - 1 $$

The $$ 9t $$ and $$ -9t $$ cancel:

$$ 8s^3 - 1 = 12t s^2 - 4t^3 - 1 $$

Adding 1 to both sides:

$$ 8s^3 = 12t s^2 - 4t^3 $$

Bringing all terms to the left:

$$ 8s^3 - 12t s^2 + 4t^3 = 0 $$

Dividing by 4 to simplify:

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

This cubic equation in $$ s $$ must be satisfied by the parameter values where the tangent intersects the curve. Since P corresponds to $$ s = t $$, $$ (s - t) $$ is a factor. Factoring the cubic:

Using synthetic division with root $$ t $$:

Coefficients: $$ 2 $$ (for $$ s^3 $$), $$ -3t $$ (for $$ s^2 $$), $$ 0 $$ (for $$ s $$), $$ t^3 $$ (constant term).

Synthetic division:

Bring down 2, multiply by $$ t $$: $$ 2t $$, add to next coefficient: $$ -3t + 2t = -t $$, multiply by $$ t $$: $$ -t^2 $$, add to next coefficient: $$ 0 + (-t^2) = -t^2 $$, multiply by $$ t $$: $$ -t^3 $$, add to last coefficient: $$ t^3 + (-t^3) = 0 $$.

Quotient: $$ 2s^2 - t s - t^2 $$, remainder 0. Thus,

$$ 2s^3 - 3t s^2 + t^3 = (s - t)(2s^2 - t s - t^2) $$

Setting to zero:

$$ (s - t)(2s^2 - t s - t^2) = 0 $$

So, $$ s - t = 0 $$ or $$ 2s^2 - t s - t^2 = 0 $$.

The solution $$ s = t $$ corresponds to point P. Solving the quadratic equation:

$$ s = \frac{t \pm \sqrt{(-t)^2 - 4 \cdot 2 \cdot (-t^2)}}{4} = \frac{t \pm \sqrt{t^2 + 8t^2}}{4} = \frac{t \pm \sqrt{9t^2}}{4} = \frac{t \pm 3t}{4} $$

Thus, $$ s = \frac{t + 3t}{4} = t $$ or $$ s = \frac{t - 3t}{4} = -\frac{t}{2} $$.

The solution $$ s = t $$ is point P, so Q corresponds to $$ s = -\frac{t}{2} $$.

Now, find the coordinates of Q:

$$ x = 4\left(-\frac{t}{2}\right)^2 + 3 = 4 \cdot \frac{t^2}{4} + 3 = t^2 + 3 $$

$$ y = 8\left(-\frac{t}{2}\right)^3 - 1 = 8 \cdot \left(-\frac{t^3}{8}\right) - 1 = -t^3 - 1 $$

Therefore, the coordinates of Q are $$ (t^2 + 3, -t^3 - 1) $$.

Comparing with the options:

A. $$ (16t^2 + 3, -64t^3 - 1) $$

B. $$ (4t^2 + 3, -8t^3 - 1) $$

C. $$ (t^2 + 3, t^3 - 1) $$

D. $$ (t^2 + 3, -t^3 - 1) $$

Option D matches the coordinates $$ (t^2 + 3, -t^3 - 1) $$. Note that for $$ t = 0 $$, the tangent does not intersect the curve again, but the problem implies $$ t \neq 0 $$ as there is another intersection point Q.

Hence, the correct answer is Option D.