Reflection of ray of light on parabolic mirror

Question

A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror whose equation is$ (y – 3)^2 = 8(x + 1)$. After reflection, the ray must pass through the point

(A) (1, 3)

(B) (–1, 3)

(C) (1, –3)

(D) (–1, –3)

My approach: Let the ray of light is y=c

As it is a mirror it will reflect at $90^o$ but I am not able to proceed from here

Answer 1

I'll give you the approach. You know the equation of parabolic mirror. The meeting point of this mirror with ray will be $(\frac{(c-3)^2}{8}-1,c)$ .

You can find the normal of the parabola at this point which is $\frac{-dx}{dy}$ . The angle that this ray $y=c$ makes with the normal is the angle the reflected ray will make with the normal. (law of reflection)

So, you have the slope of reflected line and you know it passes through $(\frac{(c-3)^2}{8}-1,c)$. So, you can find the equation of the reflected line.

Answer 2

You can proceed by a process of elimination.

The parabola’s axis is clearly parallel to the $x$-axis, and you can read from its equation that its vertex is at $(-1,3)$. You’re looking for a point through which the reflection of every ray parallel to the axis passes. So, the ray that reflects off the vertex must also pass through through this point, but this ray is just reflected back onto itself, so the answer must be either A or B.

For any $m\ne0$, there is a point on the parabola at which the tangent to it has slope equal to $m$. In particular, there is a point which is not the vertex at which the tangent makes a 45-degree angle with the $x$-axis, hence so does the normal to the parabola at that point. The reflection of an incoming horizontal ray that strikes the mirror at this point is vertical, but this reflected ray then can’t pass through the parabola’s vertex, which leaves A as the only possibility. This happens to be the parabola’s focus, as noted in a comment to your question.