## Friday, 27 May 2011

### A Geome-Treat with a Calculus Twist

So you start with a circle, let’s use x^2 + y^2 = 25 as a specimen.   We find a point on the curve, say (3,4), and decide we want to find the line tangent to the circle at that point.

For me it is(was) always a rush showing kids after a year of calculus that it can be done by just thinking of the equation as x(x) + y(y) = 25 and then replacing the x and y in parentheses with the coordinates 3 and 4.  The line 3x +4y = 25 not only passes through the point (3,4), but it is also tangent to the circle.

Ok, so suppose we do that again.. maybe we pick (4,-3) and write the line 4x-3y = 25.
Now we have two lines.
Perhaps we take the time to figure out that the two lines must intersect at (7,1) and in passing we wonder what happens if we plug that point into the x(x) + y(y) = 25 the same way? We look at the result, 7x+y=25, and decide that this time it not only doesn’t go through the point (7,1), it  certainly will not be tangent.
But what the heck. We have come this far so we can graph that line, too…

Ok, so the treat here is that you can do this with ANY conic; ellipses, hyperbolas, circles or parabolas.   If you have a point on the curve, you can find the tangent line with a simple substitution.  And if you have a point not on the conic and want to draw the tangents to the curve through that point, you just plug in that point and it gives you the line through the two tangents.

Parabolas can be a little trickier so here is an example of finding the two tangents through a point not on the curve.
I’ll use the simple y=x2 for the parabola, and pick a point (1,-1) for the exterior point.  We can replace one of the x variables in y= xx with the 1, but what about the y value.  Since there is only the one y, we can replace it with the average of y and -1, or (y-1)/2.  That gives us the line y-1 =2x , which is just y= 2x+1.  When we graph the whole thing we see that, indeed, the line cuts through points from which tangents to the curve would pass through our given point.

But, of course, the calculus student must now show that it always works.
Shall we make that due Wednesday?