Tangents drawn from the point (-8, 0) to the parabola $$y^2 = 8x$$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to:

We are given the parabola $$y^{2}=8x$$. A standard result is that any parabola of the form $$y^{2}=4ax$$ has its focus at $$(a,0)$$. Here, $$4a=8$$, so $$a=2$$ and therefore the focus is

$$F(2,0).$$

For the same parabola, a convenient parametric representation of points on the curve is

$$\bigl(at^{2},\,2at\bigr),$$

where $$t$$ is a real parameter. Substituting $$a=2$$ gives every point on the curve as

$$\bigl(2t^{2},\,4t\bigr).$$

A standard formula for the tangent to $$y^{2}=4ax$$ at the parametric point $$(at^{2},2at)$$ is

$$ty = x + at^{2}.$$

Putting $$a=2$$, the tangent at the point corresponding to the parameter $$t$$ is

$$ty = x + 2t^{2}. \quad -(1)$$

We are told that this tangent must pass through the external point $$(-8,0)$$. Hence we substitute $$x=-8$$ and $$y=0$$ into (1):

$$t\cdot 0 = -8 + 2t^{2}.$$

This simplifies step by step as

$$0 = -8 + 2t^{2},$$

$$2t^{2}=8,$$

$$t^{2}=4,$$

$$t=\pm 2.$$

Thus the two tangents from $$(-8,0)$$ touch the parabola at the parameters

$$t_{1}=2,\qquad t_{2}=-2.$$

Let $$P$$ correspond to $$t=2$$ and $$Q$$ to $$t=-2$$. Using the parametric form $$(2t^{2},4t)$$ we obtain

For $$t=2$$:

$$P\;=\;\bigl(2(2)^{2},\,4(2)\bigr)=\bigl(2\cdot 4,\,8\bigr)=(8,8).$$

For $$t=-2$$:

$$Q\;=\;\bigl(2(-2)^{2},\,4(-2)\bigr)=\bigl(2\cdot 4,\,-8\bigr)=(8,-8).$$

We now have the three vertices of the triangle:

$$P(8,8),\quad F(2,0),\quad Q(8,-8).$$

To find the area of triangle $$PFQ$$ we use the determinant (shoelace) formula:

For points $$(x_{1},y_{1}), (x_{2},y_{2}), (x_{3},y_{3}),$$ the area is

$$\text{Area}=\dfrac12\left|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right|.$$

Assigning $$(x_{1},y_{1})=(8,8),\;(x_{2},y_{2})=(2,0),\;(x_{3},y_{3})=(8,-8),$$ we substitute:

$$\begin{aligned} \text{Area}&=\dfrac12\Bigl|\,8\bigl(0-(-8)\bigr)+2\bigl((-8)-8\bigr)+8\bigl(8-0\bigr)\Bigr|\\[4pt] &=\dfrac12\Bigl|\,8(8)+2(-16)+8(8)\Bigr|\\[4pt] &=\dfrac12\Bigl|\,64-32+64\Bigr|\\[4pt] &=\dfrac12\bigl|\,96\bigr|\\[4pt] &=\dfrac{96}{2}=48. \end{aligned}$$

Hence, the correct answer is Option A.