## Tuesday, September 2, 2008

### Setting the Tone in Precalculus - Another Coordinate Investigation

Note: Read the first comment I posted which suggests a purely Euclidean geometry approach to this problem...

Don't forget our MathAnagram for Aug-Sept. Thus far we have received a couple of correct responses. You are encouraged to make a conjecture!
Look here for directions. Here is the anagram again:

PRINCE? NAH! E-ROI!

Tangent problems are usually the domain of calculus but we can keep them within the reach of geometry and algebra if we restrict our attention to circles. The calculus student spends a considerable amount of time solving a wide variety of "tangent to the curve" exercises. As any calculus instructor will tell you, many of the harder problems ask students to determine the equations of the tangent to a curve from a point not on the curve. The issue there is not the calculus. It's all about an understanding of the interface between the algebra and geometry, the essence of coordinate methods. I developed this investigation specifically to address this issue before students enter calculus. Might be another "fun" problem to start the year off with. If nothing else, it will establish the rigor of your precalculus course early on!

Part I of Investigation
Determine the coordinates of the points of tangency for the tangent lines to the unit circle from the point (0,2).

Note: Unit circle refers to the circle of radius 1, center (0,0).

The remaining parts will all refer to this same circle.

Part II of Investigation
Repeat part I for (0,3) and (0,4).

Part III of Investigation
Show that the y-coordinate of the points of tangency for the tangent lines to the unit circle from the point (0,k) is 1/k, where k ≥ 1.

(1) The result of Part III suggests that as k increases, the y-coordinate of the point of tangency decreases (inverse ratio). Ask students what happens as k approaches 1.
Students should make sense of this visually by sketching tangent lines from various points on the y-axis above the circle.
(2) There are several effective methods for solving the above parts, however, one needs to know the fundamental relationship between a tangent line and the radius drawn to the point of tangency. From that point on, one can represent the slope in two ways or represent the y-coordinate of the point of tangent in two ways. This requires strong understanding of coordinates, graphs and algebraic relationships. You may find other methods -- share them! BTW, one could also use trig methods.
(3) I chose the unit circle and a point on the y-axis for simplicity so that the student could focus on essential ideas. However, one could generalize the result to any circle and any point outside. Have fun with that!
(4) Anyone mildly surprised by the reciprocal relationship between the y-coordinate of the point on the y-axis and the y-coordinate of the point of tangency? Can anyone make sense of that?

Dave Marain said...

Guess I'll be the first to comment! There is a purely geometric approach to proving the reciprocal relationship, using only similar triangles. See if your students can find it!

Totally_clueless said...

Seems to be simpler to just choose the tangential point(x,y) to satisfy the equation of the circle (is the assumption that these students will not know this?), and use Pythogoras to directly obtain y=1/k.

Would you classify this as an algebraic-geometric approach rather than purely geometric?

TC

Dave Marain said...

tc--
It is possible to prove it without any coordinates, i.e., a purely synthetic approach. Consider the right triangle formed by the center of the circle, the tangent segment and the segment joining the center to the point (0,k). Apply one of the altitude on hypotenuse theorems! No coordinates, only lengths are used.

However, I designed this problem for a coordinate approach.

Eric Jablow said...

Let X be the point (k, 0). Let T be the point of tangency. The x-axis strikes the circle at points A (1, 0) and B (-1, 0). Use the theorem that XA XB = XT^2.

This gives the dimensions of the right triangle OTX. Now compute its area in two different ways.

Dave Marain said...

Nice area solution, Eric. Students have difficulty remember htis approach as well as the secant-tangent power theorem for circles.
BTW, did you intentionally reverse the coordinates? In fact, the x-intercept of the tangent line described in the problem (using (0,2)) is 2/√3.

Eric Jablow said...