Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

We consider the standard form of an ellipse whose centre is at the origin and whose major axis lies along the x-axis. In such a case the equation is

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b>0.$$

The distance between the foci is $$2c$$ where, by definition of an ellipse, we have $$c^{2}=a^{2}-b^{2}.$$ The length of the minor axis is $$2b,$$ while the length of the (transverse) latus rectum is given by the standard formula

$$\text{Length of latus rectum}= \frac{2b^{2}}{a}.$$

According to the statement of the problem, the latus-rectum length is $$8$$ and the distance between the foci equals the length of the minor axis. Translating both facts into algebra, we write

$$\frac{2b^{2}}{a}=8 \quad\text{and}\quad 2c=2b.$$

The second equality simplifies immediately to

$$c=b.$$

But from the definition $$c^{2}=a^{2}-b^{2},$$ replacing $$c$$ by $$b$$ gives

$$b^{2}=a^{2}-b^{2}\quad\Longrightarrow\quad 2b^{2}=a^{2}.$$

Thus

$$a^{2}=2b^{2}\quad\Longrightarrow\quad a=\sqrt{2}\,b.$$

Now we use the first given condition, namely $$\dfrac{2b^{2}}{a}=8.$$ Substituting $$a=\sqrt{2}\,b$$ yields

$$\frac{2b^{2}}{\sqrt{2}\,b}=8.$$

Carrying out the division in the numerator and denominator,

$$\frac{2b^{2}}{\sqrt{2}\,b}=\frac{2b}{\sqrt{2}}=\frac{2b\sqrt{2}}{2}=b\sqrt{2},$$

so we have

$$b\sqrt{2}=8\quad\Longrightarrow\quad b=\frac{8}{\sqrt{2}}=4\sqrt{2}.$$

Squaring this value gives $$b^{2}=(4\sqrt{2})^{2}=16\cdot2=32.$$ Since $$a^{2}=2b^{2},$$ we obtain

$$a^{2}=2\cdot32=64\quad\Longrightarrow\quad a=8.$$

Therefore the explicit equation of the ellipse is

$$\frac{x^{2}}{64}+\frac{y^{2}}{32}=1.$$

To find which option lies on this ellipse, we substitute each point into the left side of the equation and see whether the value equals $$1.$$ Let us test each candidate:

Option A: $$(4\sqrt{2},\,2\sqrt{2}).$$ We have $$x^{2}= (4\sqrt{2})^{2}=32,\; y^{2}=(2\sqrt{2})^{2}=8.$$ Hence $$\frac{x^{2}}{64}+\frac{y^{2}}{32}= \frac{32}{64}+\frac{8}{32}=0.5+0.25=0.75\lt1,$$ so the point is inside the ellipse, not on it.

Option B: $$(4\sqrt{3},\,2\sqrt{2}).$$ Now $$x^{2}= (4\sqrt{3})^{2}=48,\; y^{2}=(2\sqrt{2})^{2}=8.$$ Thus $$\frac{x^{2}}{64}+\frac{y^{2}}{32}= \frac{48}{64}+\frac{8}{32}=0.75+0.25=1.$$ Because the sum equals $$1,$$ this point satisfies the ellipse’s equation exactly and therefore lies on the curve.

Option C: $$(4\sqrt{3},\,2\sqrt{3}).$$ Here $$x^{2}=48,\; y^{2}=(2\sqrt{3})^{2}=12.$$ Then $$\frac{x^{2}}{64}+\frac{y^{2}}{32}= \frac{48}{64}+\frac{12}{32}=0.75+0.375=1.125\gt1,$$ so the point is outside the ellipse.

Option D: $$(4\sqrt{2},\,2\sqrt{3}).$$ Now $$x^{2}=32,\; y^{2}=12,$$ giving $$\frac{x^{2}}{64}+\frac{y^{2}}{32}=0.5+0.375=0.875\lt1,$$ again inside the ellipse.

Only Option B satisfies the equation precisely.

Hence, the correct answer is Option B.