A Geome-Treat with a Calculus Twist-extended

 So you start with a circle, let’s use x^2 + y^2 = 25 as a specimen.   We pick 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(x1) + y(y1) = 25 and then replacing the x1 and y1 in parentheses with the coordinates of a point on the circle, 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…

I remind you that you probably found those same two points using a compass and straight edge in HS geometry.......

Given a circle and a point outside the circumference, you can find the two points where it will be tangent by construction of another circle, centered on the midpoint of a segment between the given point and the center of the circle.  

Ok, so the treat here is that you can do the Algebraic method above in the creation of the equation through a point on  ANY conic; ellipses, hyperbolas, circles or parabolas.   If you have a point on the curve, you can find the tangent line with the same 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. 

In along the way, we'll show a use of the extra circle method with a compass and straightedge creation of tangents from a point not on the curve,,,,except for the hyperbolas.

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?

For the geometric construction we let the line perpendicular to the axis of symmetry passing through the vertex.  (That is our extra circle in this case, it is just infinite.  We also need the focus of the parabola, (1/4,0)

Our second circle will be be centered on the midpoint of the line joining the given point, A,and the focus.  The intersections of this circle with our "infinite" circle (Points C and D) produces the second points of the two tangents on the parabola from the given point, A.

Ellipse

The algebraic approach in an ellipse works exactly like the circle and parabola.  We create an ellipse with deliberate intent to provide a surprise reentry of a previous method in the geometric solution.  x^2/25 + y^2/  16 = 1.  We calculated a point with x=1 by entering that into the equation of the ellipse and solving for y to get the point (1, 8*(6^.5) /5), or appx. (1,3.9191835...).

inserting these for one x and one y in the squared terms (1x/25+ (8*(6^.5) /5) y/16=1) we get the line y = -1/5 sqrt(2/3) (x - 25).  we can find another point on the tangent by substituting 10 in for x and solving for y, and got y=sqrt(6).  

Next we see what happens when we place these values in for one x and one y in the original ellipse.

10 x /25 + sqrt(6)y/16=1  and finding the line again has a second intersection with the original ellipse at appx (3.57, -2.8) .  

At this point I would like to return to the compass and straightedge construction of two points of tangency from a point not on the ellipse, and yes, there are two circles involved again.  We will use the same ellipse,  x^2/25 + y^2/  16 =1 and select as our external point the point (10,3), just to be near the point of the algebraic method.

We first create a circle that passes through both ends of the major axis centered at the center of the ellipse.  Then a second circle with a center that is the midpoint of the segment from our external point, D=(10,3), to the focus of the ellipse....which focus??? It turns out it doesn't matter. 

Then we find the two points of intersection from our exocircle and the point-focus circle, (G and H). The lines passing through DG and DH will be tangent to the ellipse at points J and K. So very close to the HS geometry method for a circle, how could you forget it now????