In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $$(0, 5\sqrt{3})$$, then the length of its latus rectum is:

We are told that the ellipse is centred at the origin, and that one focus is at the point $$(0,\,5\sqrt{3})$$. Because this focus lies on the $$y$$-axis, the major axis must also be along the $$y$$-axis. Hence we may write the standard form of the ellipse as

$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,$$

where $$a$$ is the semi-major axis, $$b$$ is the semi-minor axis and, by definition of “major”, we always have $$a \gt b$$.

For an ellipse in this orientation, the foci are located at $$(0,\pm c)$$, where the focal distance $$c$$ satisfies the well-known relation

$$c^{2}=a^{2}-b^{2}.$$

From the question, one of the foci is $$(0,5\sqrt{3})$$, so we immediately read off

$$c = 5\sqrt{3}.$$

Therefore, from the focal relation we have

$$a^{2}-b^{2}=c^{2}= \bigl(5\sqrt{3}\bigr)^{2}=25\cdot3=75.$$

Next, the question states that the difference of the lengths of the major and minor axes is $$10$$. The length of the major axis is $$2a$$ and that of the minor axis is $$2b$$, so

$$2a-2b=10 \quad\Longrightarrow\quad a-b=5.$$

We now possess the two equations

$$\begin{cases} a-b = 5,\\[4pt] a^{2}-b^{2} = 75. \end{cases}$$

Notice that $$a^{2}-b^{2}$$ factors conveniently:

$$a^{2}-b^{2}=(a-b)(a+b).$$

Substituting $$a-b=5$$ into this factorisation gives

$$5\,(a+b)=75 \quad\Longrightarrow\quad a+b = \frac{75}{5}=15.$$

We now have the simple linear system

$$\begin{cases} a-b = 5,\\ a+b = 15. \end{cases}$$

Adding the two equations:

$$2a = 20 \quad\Longrightarrow\quad a = 10.$$

Substituting $$a=10$$ into $$a-b=5$$ yields

$$10-b=5 \quad\Longrightarrow\quad b = 5.$$

With $$a=10$$ and $$b=5$$ found, we now compute the required length of the latus rectum. For an ellipse, the standard formula for the length $$L$$ of the latus rectum is

$$L = \frac{2b^{2}}{a}.$$

Substituting the values just obtained:

$$L = \frac{2\,(5)^{2}}{10}= \frac{2\cdot25}{10}= \frac{50}{10}=5.$$

Hence, the correct answer is Option D.