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Parabolic Mirrors and Telescopes

Date: 06/03/97 at 00:45:09
From: Derek Miyasaki
Subject: Calculus, parabolas...

Any suggestions (especially parts e and f)?

Why Astronomers Use Telescopes with Parabolic Mirrors

The star nearest our sun is Alpha Centauri, which is about 4 light 
years from earth.  Alpha Centauri is so far away that when its light 
reaches earth, it is traveling in essentially parallel rays.  To 
observe distant stars, astronomers use mirrors shaped like 
paraboloids, which are parabolas rotated about their axes.  The reason 
they use a paraboloidal mirror is that it focuses all the light to a 
single point, the focus.  This point is the image of the star in the 
paraboloidal mirror.  In this project you will demonstrate the 
focusing property of parabolas.

a) Suppose our mirror is shaped like the parabola y = kx^2, where k is 
any positive constant.  Find the coordinates of its focus and the 
equation of its directrix in terms of k.

*Definition* The line tangent to the curve with equation y = f(x) at 
a point (a,f(a)) on the curve is the line tangent through (a,f(a)) 
with slope f'(a). We call this line a tangent line.

b) Find the equation of the line tangent to the parabola at a point 
(x1,y1) on the parabola.  Then find the y-intercept of the tangent 

c) Consider the triangle formed by a point (x1,y1) on the parabola, 
the y-intercept of the tangent line at this point, and the focus. Draw 
a picture of this triangle. Prove this triangle is isosceles.

d) Suppose an incoming light ray strikes a curve at a point (x1,y1). 
If the light ray makes an angle "z" with respect to the tangent line, 
then it is reflected at an equal angle to the tangent line. This 
result from physics is known by the phrase "the angle of incidence 
equals the angle of reflection."  Using this fact, argue that incoming 
light rays parallel to the axis of the parabola are all reflected to 
the focus, independent of the point of incidence.  Thus, a parabolic 
mirror focuses incoming light rays parallel to the axis to a point.

e) The path followed by a ray of light from the star to the focus of 
the mirror has another special property.  Draw a chord of the parabola 
that is above the focus and parallel to the directrix.  Consider a ray 
of light parallel to the axis as it crosses the chord, hits the 
parabola and is reflected to the focus. Let d1 be the distance from 
the chord to the point of incidence (x1,y1) on the parabola and let d2 
be the distance from (x,y) to the focus.  Show that the sum of the 
distances d1+d2 is constant, independent of the particular point of 

f) Extend your argument from the parabolic cross section to the entire 
paraboloidal mirror, obtained by rotating this cross section about the 
y-axis.  Thus, prove that all incident rays parallel to the axis of 
such a mirror focus to a point.  Also prove that light traveling along 
different rays to the focus from a chord perpendicular to the axis 
will have traveled the same distance, even though these rays were 
reflected from different points of incidence on the mirror.  Thus all 
of the light waves will arrive in phase, and so interfere 
constructively to produce a nice bright spot as the image of the star.

Date: 06/03/97 at 12:01:42
From: Doctor Anthony
Subject: Re: Calculus, parabolas...


a) The standard equation of a parabola is y^2 = 4ax. With this 
equation the focus is at (a,0) and the directrix is the line x = -a.

We convert the equation y = kx^2 into this form if we write x^2 = 4ay. 
Then the focus would be at (0,a) and directrix would be y = -a.

But x^2 = (1/k)y, giving 4a = 1/k, so a = 1/(4k).

Thus the focus is (0,1/(4k)) and the directrix is the line 
y = -1/(4k).

b) y = kx^2, so dy/dx = 2kx and the equation of the tangent at (x1,y1) 
     y-y1 = 2kx1(x-x1)       This cuts the y-axis where x = 0
     y-y1 = -2k(x1)^2
          = -2y1
        y = -y1   

So the tangent cuts the y-axis the same distance below the origin as 
the point (x1,y1) is above the x-axis.

c) The three points are (0,1/(4k)), (x1,y1), and (0,-y1).

The distance from the focus to the point (0,-y1) is y1+1/(4k).

The distance from the focus to (x1,y1) is:
              sqrt(x1^2 + (y1-1/(4k))^2)
            = sqrt((y1/k) + y1^2 - y1/(2k) + 1/(4k)^2)
            = sqrt(y1^2 + y1/(2k) + 1/(4k)^2)
            = sqrt(y1+1/(4k))^2
            = y1 + 1/4k

So the distances from the focus to other points are equal, hence the 
triangle is isosceles.
d) This follows from the fact that base angles of an isosceles 
triangle are equal.  The angle between the incoming ray (parallel to 
the y-axis) and the tangent is angle of incidence, and this equals the 
angle between the tangent and the line to the focus from (x1,y1) (the 
angle of reflection). So every ray parallel to the y-axis will be 
reflected to pass through the focus.

e) If the chord is the line y = c, then the distance from the point 
where this chord cuts the parabola to the focus is  c + 1/(4k) (proved 

The distance d1 from y = c to y = y1 is c - y1, and so the distance 
covered by the ray which is incident on the parabola at (x1,y1) is 
c-y1 + y1 + 1/(4k) = c + 1/(4k), which is the same as for the other 

This would apply to any other ray incident, say at (x2,y2), and shows 
that all rays crossing the chord have an equal path length to the 

f) If we cut the parabolic dish by a plane through its axis, the 
arguments given above apply to the incoming rays lying in the plane.  
So all rays in all such planes will meet at the focus.

-Doctor Anthony,  The Math Forum
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