Thursday, May 15, 2008

When Curves Collide Part II - Quadratic Systems Re-Explored!

One of MathNotations more popular posts (hundreds of views) was published one year ago this week: When Curves Collide.

Here's a variation to review the essential ideas or to use as an assessment problem or just to challenge yourself. Parts (a) thru (d) require some theoretical analysis and algebraic skill. Part (e) is the main challenge...

An Investigation for Algebra 2/Precalculus
Consider the quadratic-quadratic system:

x2 + y2 = 1
y = ax2 -1, a>0

(a) Show that (0,-1) is always a solution to this system.

(b) For what values of the parameter 'a' will there be 3 distinct solutions to the system?
Coordinate Interpretation: For what values of 'a' will the parabola and circle intersect in 3 distinct points?

(c) For what value(s) of the parameter 'a' will two of the points of intersection be above the x-axis? Below the x-axis (in addition to (0,-1))? On the x-axis?

(d) For the case that there are 3 distinct solutions, determine the two solutions, other than (0,-1), in terms of 'a'.

(e) Now for the main problem:

Assume the graph of our system has three points of intersection: P, Q and R(0,-1). If the area of ΔPQR is 32/25, determine the coordinates of P and Q and the value of 'a'.

(f) Can you think of an even more clever variation!


Florian said...

Just a quick shot at the main
problem (it seems alot easier
than the preceding part):

Since the parabola is symmetrical
and R(0,-1) the point in the line
of the symmetry the triangle is
split by the y-Axis into two
equilateral trianlgles.

This means we can simply calculate
the area of the rectangle adjacent
to the y-axis where P and R are
two diagonal opposite points.

=> Px*Ry = 32/25 <=> Rx = -32/25 = -Qx

Then plugin Rx into the circle
equation and get Ry. Qy = Ry
because of the symmetry. Thats all.

Dave Marain said...

That's pretty much it, Florian. However, work out the details and tell me if you notice anything about the numerical result.

Also, if one does the preceding parts (and you're right, they may be more difficult!), you would have most of the algebra done which can then be applied easily to the area.

Actually, this investigation has many purposes, perhaps the most important of which is to develop the concept of a parameter. I believe this is not adequately developed in most courses leading up to calculus. Agreed?

Eric Jablow said...

You may want the calculus interpretation too. Take the equation x²+y²=1 near (0, -1). Let's write y as a function of x:

y = -√(1-x²).

Let's take a couple of terms of the power series for the the expression in x. Find the Maclaurin series for√(1+x) and replace x with -x²:

y = -1 + 1/2 x² + 1/8 x⁴ + ...

For x small enough, this behaves like its first two terms. So, consider the parabola

y = -1 + ax².

If a > 1/2, for x close enough to 0, the parabola is higher than the circle arc near (0, -1). So, the parabola enters the interior of the circle in both directions. It has to leave, and it does at two places, for a total of three.

Incidentally, both the circle and the parabola are conics, quadratic curves. Two conics should meet in 4 points, according to Bezout's theorem, although this may require extending to complex numbers (and would if a < 1/2). Why only three points then? Because the two curves meet at (0, -1) 'doubly'; the difference between the two expressions for y doesn't involve linear terms in x.

Dave Marain said...

Ah, the "power of series" to give insight to many a problem, Eric! I would love to see more precalculus and calculus students learn to appreciate how Newton generalized the binomial formula to rational exponents as in (1+x)^(1/2). This background helps them to appreciate Taylor's and Maclaurin's Series.

In my next post, I will publish the answers to the questions posed in this post and i will include a fairly easy circle problem that developed naturally for me from the current problem. Wouldn't it be wonderful if our students could learn, in a math classroom, this most important idea of research:

As we search for a solution to one problem, we discover many other relationships and important mathematics along the way. How much extraordinary mathematics was discovered serendipitously on the road to cracking Fermat's Last Theorem!