History of Math: Conics

[warwick]

This is from my History of Mathematics course.

Prop. I-34 states: Let C be a point on a hyperbola, CB be the perpendicular from that point to the diameter. Let G and H be the intersections of the diameter with the curve, and choose A on the diameter or the diameter extended so that AH:AG = BH:BG. Then AC will be tangent to the curve at C.

This result can be stated algebraically by letting AG = t and BG = x. so in the case of a hyperbola, BH = 2a+x and AH = 2a - t, therefore (2a-t)/t=(2a+x)/x. Solving for t gives t=ax/(a+x).

Well, I centered the hyperbola at the origin and found the derivative. It gives me the slope of the tangent line at whatever point I want. How do I prove this?

 

[warwick]

From the professor:

Hint for the problem in Lecture 11: Write the equation x^2/a^2 - y^2/ b^2=1 for the hyperbola. Denote H=(a, 0), G=(-a, 0), a>0, B=(-c, 0), c>0. Then find C and find the equation of the tangent line of the hyperbola at C. Find the A.

In finding C, I solved for y and since C is associated with x = -c at B, I inserted x = -c into the equation for y.

I found the derivative of y. I could simply substitute x = -c again and get the slope of the tangent line at C.

 

[warwick]

Anyone?