Consider the parabola with vertex $$\left(\frac{1}{2}, \frac{3}{4}\right)$$ and the directrix $$y = \frac{1}{2}$$. Let P be the point where the parabola meets the line $$x = -\frac{1}{2}$$. If the normal to the parabola at P intersects the parabola again at the point Q, then $$(PQ)^2$$ is equal to:

We begin with the geometric definition of a parabola: every point on the curve is equidistant from its focus and its directrix. The vertex is given as $$\left(\dfrac12,\dfrac34\right)$$ and the directrix is the horizontal line $$y=\dfrac12$$, so the axis of the parabola is vertical.

The distance from the vertex to the directrix is $$\dfrac34-\dfrac12=\dfrac14$$. This same distance, denoted by $$a$$, separates the vertex from the focus but in the opposite (upward) direction. Hence the focus is $$\left(\dfrac12,\dfrac34+\dfrac14\right)=\left(\dfrac12,1\right)$$ and we have $$a=\dfrac14$$.

For a parabola with vertical axis, vertex $$(h,k)$$ and focal length $$a$$, the standard form is $$ (x-h)^2 = 4a\,(y-k). $$ Here $$h=\dfrac12,\;k=\dfrac34,\;4a=4\left(\dfrac14\right)=1$$, so the explicit equation of the parabola is $$ (x-\dfrac12)^2 = y-\dfrac34, \qquad\text{or}\qquad y = (x-\dfrac12)^2 + \dfrac34. $$

We next locate the point $$P$$ where the parabola meets the vertical line $$x=-\dfrac12$$. Substituting $$x=-\dfrac12$$ into the equation of the parabola,

$$ y = \Bigl(-\dfrac12-\dfrac12\Bigr)^2 + \dfrac34 = (-1)^2 + \dfrac34 = 1 + \dfrac34 = \dfrac74. $$

Thus $$ P\left(-\dfrac12,\dfrac74\right). $$

To find the normal at $$P$$, we first need the slope of the tangent. Re-writing the parabola as $$y=(x-\dfrac12)^2+\dfrac34$$, we differentiate:

$$ \frac{dy}{dx}=2(x-\dfrac12). $$

At $$x=-\dfrac12$$, $$ m_{\text{tan}}=2\Bigl(-\dfrac12-\dfrac12\Bigr)=2(-1)=-2. $$ Hence the slope of the normal is the negative reciprocal: $$ m_{\text{norm}}=\dfrac12. $$

The normal through $$P(-\dfrac12,\dfrac74)$$ therefore satisfies $$ y-\dfrac74=\dfrac12\bigl(x+\dfrac12\bigr). $$ Adding $$\dfrac74$$ to both sides gives the straight-line equation $$ y=\dfrac12x+\dfrac14+\dfrac74=\dfrac12x+2. $$

Let the normal meet the parabola again at $$Q(x,y)$$. Setting the expression for $$y$$ on the line equal to the expression for $$y$$ on the parabola, we obtain

$$ (x-\dfrac12)^2+\dfrac34=\dfrac12x+2. $$

Simplifying step by step:

$$ (x-\dfrac12)^2 = \dfrac12x + 2 - \dfrac34 = \dfrac12x + \dfrac54, $$ $$ x^2 - x + \dfrac14 = \dfrac12x + \dfrac54, $$ $$ x^2 - x - \dfrac12x + \dfrac14 - \dfrac54 = 0, $$ $$ x^2 - \dfrac32x - 1 = 0. $$

Multiplying by $$2$$ to clear the fraction,

$$ 2x^2 - 3x - 2 = 0. $$

The quadratic formula, $$ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}, $$ with $$a=2,\;b=-3,\;c=-2$$ gives

$$ x=\dfrac{3\pm\sqrt{(-3)^2-4(2)(-2)}}{2\cdot2} =\dfrac{3\pm\sqrt{9+16}}{4} =\dfrac{3\pm5}{4}. $$

Thus $$x=-\dfrac12$$ or $$x=2$$. The first solution reproduces point $$P$$, so the second root corresponds to $$Q$$ with $$x=2$$. Substituting into $$y=\dfrac12x+2$$, we find

$$ y=\dfrac12\cdot2+2=1+2=3, \qquad Q(2,3). $$

Finally we compute the squared distance between $$P\bigl(-\dfrac12,\dfrac74\bigr)$$ and $$Q(2,3)$$. Using the distance formula $$ (PQ)^2=(x_2-x_1)^2+(y_2-y_1)^2, $$ we have

$$ x_2-x_1 = 2-\Bigl(-\dfrac12\Bigr)=\dfrac52, \qquad y_2-y_1 = 3-\dfrac74 = \dfrac{12}{4}-\dfrac74=\dfrac54. $$

Therefore

$$ (PQ)^2 = \left(\dfrac52\right)^2 + \left(\dfrac54\right)^2 = \dfrac{25}{4}+\dfrac{25}{16} = \dfrac{100}{16}+\dfrac{25}{16} = \dfrac{125}{16}. $$

Hence, the correct answer is Option C.