Let a parabola $$P$$ be such that its vertex and focus lie on the positive $$x$$-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $$O(0, 0)$$ to the parabola $$P$$ which meet $$P$$ at $$S$$ and $$R$$, then the area (in sq. units) of $$\triangle SOR$$ is equal to:

We are told that the vertex of the required parabola lies at a distance 2 units from the origin on the positive $$x$$-axis, so the vertex is $$V(2,0)$$. Its focus is 4 units from the origin on the same axis, i.e. at $$F(4,0)$$. Because the focus is to the right of the vertex, the parabola opens to the right, and its axis is the $$x$$-axis itself.

The standard equation of a right-opening parabola with vertex $$(h,k)$$ is

$$ (y-k)^2 = 4a\,(x-h), $$

where $$a$$ denotes the distance from the vertex to the focus. Here $$h=2,\;k=0$$ and

$$ a = VF = 4-2 = 2. $$

Substituting these values, we obtain the explicit equation of the parabola:

$$ y^2 = 4a\,(x-2) = 4\cdot 2\,(x-2) = 8(x-2). $$

Now we must draw tangents to this parabola from the origin $$O(0,0)$$. To do that, we first recall the point-form of the tangent to a parabola. For the standard parabola $$Y^2 = 4aX$$ with vertex at the origin, the tangent at a point $$(X_1,Y_1)$$ on the curve is

$$ Y\,Y_1 = 2a\,(X+X_1). $$

Our parabola is translated 2 units to the right. Let us therefore shift coordinates:

$$ X = x-2,\qquad Y = y. $$

In these new coordinates the equation becomes $$Y^2 = 8X,$$ still with $$a=2$$. Hence, at $$(X_1,Y_1)$$ (equivalently $$(x_1,y_1)$$), the tangent is

$$ Y\,Y_1 = 2a\,(X+X_1)\; \Longrightarrow\; y\,y_1 \;=\; 2\bigl[(x-2)+(x_1-2)\bigr]. $$

Simplifying the right-hand side, this gives

$$ y\,y_1 = 2\,(x + x_1 - 4). \quad -(1) $$

Because we want this tangent to pass through the origin $$O(0,0)$$, we substitute $$(x,y)=(0,0)$$ into (1):

$$ 0\cdot y_1 = 2\,(0 + x_1 - 4)\;\; \Longrightarrow\;\; 2\,(x_1 - 4)=0. $$

Therefore

$$ x_1 = 4. $$

To find the corresponding $$y_1$$-coordinates of the points of contact, we insert $$x_1=4$$ into the parabola’s equation:

$$ y_1^2 = 8\,(4-2)=8\cdot 2 = 16 \;\Longrightarrow\; y_1 = \pm 4. $$

Hence the two points of tangency are

$$ S(4,4)\quad\text{and}\quad R(4,-4). $$

We now possess the three vertices of the required triangle: $$O(0,0),\;S(4,4),\;R(4,-4).$$ Since one vertex is the origin, the area formula simplifies. For a triangle with vertices $$O(0,0),\;B(x_1,y_1),\;C(x_2,y_2)$$ the area is

$$ \text{Area} = \dfrac12\,\bigl|\,x_1y_2 - x_2y_1\,\bigr|. $$

Substituting $$B=S(4,4)$$ and $$C=R(4,-4)$$, we get

$$ \text{Area} = \frac12\,\Bigl|\,4\cdot(-4)\;-\;4\cdot4\,\Bigr| = \frac12\,\bigl|\, -16 - 16 \bigr| = \frac12\,(32) = 16. $$

Hence, the required area is $$16$$ square units.

Hence, the correct answer is Option B.