Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If $$\Delta S'BS$$ is a right angled triangle with right angle at B and area ($$\Delta S'BS$$) = 8 sq. units, then the length of a latus rectum of the ellipse is:

We denote the semi-major axis by $$a$$, the semi-minor axis by $$b$$ and the eccentricity by $$e$$. For an ellipse whose major axis is the $$x$$-axis, the foci are $$S(ae,0)$$ and $$S'(-ae,0)$$, while the extremities of the minor axis are $$B(0,b)$$ and $$B'(0,-b)$$. In the problem we take $$B(0,b)$$.

First recall the two standard relations for an ellipse:

$$b^{2}=a^{2}(1-e^{2}) \qquad\text{and}\qquad \text{length of a latus rectum}= \dfrac{2b^{2}}{a}.$$

We are told that $$\triangle S'BS$$ is right-angled at $$B$$. The vectors $$\overrightarrow{BS}=(ae, -\,b)$$ and $$\overrightarrow{BS'}=(-ae, -\,b)$$ are therefore perpendicular, so their dot product must vanish:

$$\overrightarrow{BS}\cdot\overrightarrow{BS'} = (ae)(-ae)+(-b)(-b)= -\,a^{2}e^{2}+b^{2}=0.$$

Thus we obtain

$$b^{2}=a^{2}e^{2} \quad -(1).$$

Combining (1) with the basic ellipse relation $$b^{2}=a^{2}(1-e^{2})$$ we get

$$a^{2}e^{2}=a^{2}(1-e^{2}) \;\;\Longrightarrow\;\; e^{2}=1-e^{2} \;\;\Longrightarrow\;\; e^{2}=\dfrac12,$$

and hence

$$e=\dfrac1{\sqrt2}.$$

Next, we use the fact that the triangle is right-angled at $$B$$ and its area is $$8$$ sq. units. For a right triangle the area formula is

$$\text{Area}=\dfrac12(\text{product of the perpendicular sides}).$$

The perpendicular sides here are $$BS$$ and $$BS'$$, whose lengths are equal:

$$BS=\sqrt{(ae)^{2}+b^{2}}, \quad BS'=\sqrt{(-ae)^{2}+b^{2}}=\sqrt{(ae)^{2}+b^{2}}.$$

From (1) we have $$b^{2}=a^{2}e^{2}$$, so

$$BS=\sqrt{a^{2}e^{2}+a^{2}e^{2}}=\sqrt{2a^{2}e^{2}}=a e\sqrt2.$$

Since $$e=\dfrac1{\sqrt2}$$, this simplifies to

$$BS=a.$$

Now set up the area condition:

$$\dfrac12\,(BS)\,(BS')=\dfrac12\,a\,a=8 \;\;\Longrightarrow\;\; a^{2}=16 \;\;\Longrightarrow\;\; a=4.$$

With $$a=4$$, relation (1) gives

$$b^{2}=a^{2}e^{2}=16\left(\dfrac12\right)=8.$$

Finally, insert $$a=4$$ and $$b^{2}=8$$ into the latus rectum formula:

$$\text{latus rectum}= \dfrac{2b^{2}}{a}= \dfrac{2\times 8}{4}=4.$$

Hence, the correct answer is Option C.