A ray of light through $$(2, 1)$$ is reflected at a point $$P$$ on the $$y$$-axis and then passes through the point $$(5, 3)$$. If this reflected ray is the directrix of an ellipse with eccentricity $$\frac{1}{3}$$ and the distance of the nearer focus from this directrix is $$\frac{8}{\sqrt{53}}$$, then the equation of the other directrix can be:

First we note that the light is incident from the point $$(2,1)$$, strikes a point $$P$$ on the $$y$$-axis and afterwards travels through the point $$(5,3)$$. A vertical mirror is modelled by the line $$x=0$$ (the $$y$$-axis). Whenever a ray reflects from a straight mirror, the “law of reflection” tells us that the actual broken path is equivalent to a straight line that joins the image of the source with the final point. Concretely, if we reflect the source $$A(2,1)$$ in the line $$x=0$$, its mirror image becomes $$A'(-2,1)$$. Therefore the straight line $$A'B$$, where $$B(5,3)$$ is the final point, meets the $$y$$-axis exactly at the required point of incidence $$P$$.

The slope of the line joining $$A'(-2,1)$$ and $$B(5,3)$$ is

$$m=\frac{3-1}{5-(-2)}=\frac{2}{7}.$$

Using the two-point form, the equation of $$A'B$$ is

$$y-1=\frac{2}{7}\,(x+2).$$

To find $$P$$ we put $$x=0$$ (because $$P$$ lies on the $$y$$-axis):

$$y-1=\frac{2}{7}\,(0+2)=\frac{4}{7}\;\Longrightarrow\;y=1+\frac{4}{7}=\frac{11}{7}.$$

Hence $$P\left(0,\frac{11}{7}\right)$$. After reflection the ray simply follows the same straight line through $$P$$ and $$B$$, so this reflected ray is the line

$$y-\frac{11}{7}=\frac{2}{7}\,(x-0).$$

Multiplying by $$7$$ gives $$7y-11=2x$$, or, in the standard form,

$$2x-7y+11=0.$$

This line is given in the statement to be a directrix of an ellipse whose eccentricity is $$e=\dfrac13$$.

For any ellipse the two directrices are parallel; the major axis is perpendicular to them. Every focus lies on this axis. The distance from a focus to its corresponding directrix is related to the semi-major axis length $$a$$ by the well-known formula

$$\text{distance (focus, directrix)} = a\!\left(\frac1e-e\right).$$

We are told that the nearer focus is at a perpendicular distance $$\dfrac{8}{\sqrt{53}}$$ from the directrix $$2x-7y+11=0$$. Therefore, writing the above relation explicitly, we have

$$a\!\left(\frac1e-e\right)=\frac{8}{\sqrt{53}}.$$

Substituting $$e=\dfrac13$$ gives

$$a\!\left(3-\frac13\right)=a\!\left(\frac{9-1}{3}\right)=a\!\left(\frac{8}{3}\right)=\frac{8}{\sqrt{53}},$$ $$\Longrightarrow\;a=\frac{8}{\sqrt{53}}\cdot\frac{3}{8}=\frac{3}{\sqrt{53}}.$$

For an ellipse the distance between its two directrices is $$\dfrac{2a}{e}$$ (because each directrix is at the distance $$\dfrac{a}{e}$$ from the centre on opposite sides). Hence, with $$a=\dfrac{3}{\sqrt{53}}$$ and $$e=\dfrac13$$, the separation of the directrices is

$$\frac{2a}{e}=2a\!\left(\frac{1}{e}\right)=2a\cdot3=6a=6\left(\frac{3}{\sqrt{53}}\right)=\frac{18}{\sqrt{53}}.$$

Now let us write the equation of a general line that is parallel to the given directrix. Any line parallel to

$$2x-7y+11=0$$

has the form

$$2x-7y+k=0,$$

where $$k$$ is a constant. The perpendicular distance between two such parallel lines

$$2x-7y+11=0\quad\text{and}\quad 2x-7y+k=0$$

is

$$\frac{|k-11|}{\sqrt{2^{2}+(-7)^{2}}}=\frac{|k-11|}{\sqrt{4+49}}=\frac{|k-11|}{\sqrt{53}}.$$

Setting this equal to the required separation $$\dfrac{18}{\sqrt{53}}$$, we obtain

$$\frac{|k-11|}{\sqrt{53}}=\frac{18}{\sqrt{53}} \;\Longrightarrow\; |k-11|=18.$$

So

$$k-11=18\quad\text{or}\quad k-11=-18,$$ $$k=29\quad\text{or}\quad k=-7.$$

Consequently the second directrix can be either

$$2x-7y+29=0\quad\text{or}\quad 2x-7y-7=0.$$

These two equations appear exactly in option (C).

Hence, the correct answer is Option C.