Finding Tangent Lines on a Hyperbola: Proving Proposition I-34

[Shackleford]

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?

 

[Shackleford]

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.

 

[dynamicsolo]

From the professor:


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.

Do you know implicit differentiation? If you multiply the equation for the hyperbola through by a²b² and then differentiate both sides with respect to x , you will get an expression for dy/dx in terms of x and y . (You could also solve the equation of the hyperbola for y and get an explicit form for y' as a function of x , but it's not as tidy...)

You now have the slope of the tangent line to any point (x,y) on the hyperbola (and you can get y as a function of x from the equation of the hyperbola, but we can save that for later -- and may not even need it). Write a "point-slope" equation for the tangent line passing through (x,y) on the hyperbola and find its x-intercept (what is x for y=0?). That's where point A is. You could now check the requested proportions...

 

[Shackleford]

Do you know implicit differentiation? If you multiply the equation for the hyperbola through by a²b² and then differentiate both sides with respect to x , you will get an expression for dy/dx in terms of x and y . (You could also solve the equation of the hyperbola for y and get an explicit form for y' as a function of x , but it's not as tidy...)

You now have the slope of the tangent line to any point (x,y) on the hyperbola (and you can get y as a function of x from the equation of the hyperbola, but we can save that for later -- and may not even need it). Write a "point-slope" equation for the tangent line passing through (x,y) on the hyperbola and find its x-intercept (what is x for y=0?). That's where point A is. You could now check the requested proportions...

I haven't used implicit differentiation since Cal 1. I did find the explicit form for y', and it's not very tidy at all. Let me try this way.

b²2x - a²2yy' = 0

y' = [b²/a²][x/y]

I still need to find C. If I use the explicit form for y, it's not pretty.

C = (-c, [1/a²][bc²-a²b²]^1/2)

(y-y₁) = m(x-x₁)

(y-y₁) = [b²/a²][x/y](x-x₁)

 

[dynamicsolo]

(y-y₁) = m(x-x₁)

(y-y₁) = [b²/a²][x/y](x-x₁)

This looks like what I have. If you are taking C( x₁, y₁ ) to be the point on the hyperbola at which you are constructing the tangent line, then the slope there is
m = [b²/a²][x₁/y₁]and the equation for the tangent line is

(y - y₁) = [b²/a²][x₁/y₁](x-x₁) .

The x-intercept, which is the point A( X, 0 ) , would be found from setting y = 0 in that equation, or(0 - y₁) = [b²/a²][x₁/y₁](X-x₁) --> X = x₁ - [ (a²/b²) · (y₁/x₁) ]·y₁ .

Since you are working on a proportionality theorem, this form will probably be more useful than that which would be obtained from the explicit form for y' .

 

[Shackleford]

I need to use this notation:

H=(a, 0),
G=(-a, 0), a>0,
B=(-c, 0), c>0

The orthogonal line goes through B (x=-c) and intersects the hyperbola at C (-c,y₁).

I guess to keep my notation consistent x₁ = -c.

Using my notation, I have x = -c - [a²/b²]*c*y₁² which is the same as yours.

I'm not seeing an a clear relationship among the ratio of the lengths of the segments.

AH:AG = BH:BG

BH:BG = (a+c)/(c-a) or (2a+BG)/BG

AH:AG = [c+a+(a²/b²)*c*y₁²)] / [a-c-(a²/b²)]

Or, I fumbled around and got

(a-x₁)/-(a-x₁)

(a-x₂)/-(a-x₂).

 

[dynamicsolo]

I finally managed to get this to work, and I think I have a result akin to yours, but you have to be very careful about signs. I modified my derivation to agree with the notation in the problem.

Following the labeling back in post #1, we wish to show that

[itex] \frac{AH}{AG} = \frac{2a - t}{t} = \frac{BH}{BG} = \frac{2a + x}{x} , [/itex] or [itex] \frac{2a}{t} - 1 = \frac{2a}{x} + 1 [/itex].We will call the point on the hyperbola C( -c , Y ) . As we discussed earlier (posts #4 & 5), using implicit differentiation, the slope of the tangent line at C is [itex]m = ( \frac{b^{2}}{a^{2}} ) \cdot ( \frac{-c}{Y} ) [/itex] , with c > 0. The equation of the tangent line to C is then

[tex]( y - Y ) = ( \frac{b^{2}}{a^{2}} ) \cdot ( \frac{-c}{Y} ) \cdot ( x + c ) , [/tex]

so the x-intercept of this line is A( X, 0 ) , with X given by

[tex] ( 0 - Y ) = ( \frac{b^{2}}{a^{2}} ) \cdot ( \frac{-c}{Y} ) \cdot ( X + c ) \Rightarrow X + c = ( \frac{a^{2}}{b^{2}} ) \cdot ( \frac{Y}{c} ) \cdot Y \Rightarrow X = -c + ( \frac{a^{2}}{b^{2}c} \cdot Y^{2}) . [/tex]

But, from the equation of the hyperbola, [itex]\frac{c^{2}}{a^{2}} - \frac{Y^{2}}{b^{2}} = 1 \Rightarrow \frac{Y^{2}}{b^{2}} = \frac{c^{2}}{a^{2}} - 1 .[/itex]

Hence, [itex] X = -c + ( \frac{a^{2}}{c} ) \cdot (\frac{Y^{2}}{b^{2}} ) = -c + ( \frac{a^{2}}{c} ) \cdot ( \frac{c^{2}}{a^{2}} - 1 ) = -c + c - ( \frac{a^{2}}{c} ) = -\frac{a^{2}}{c} . [/itex] (This checks out: the coordinate X is negative.)Now, from our earlier definitions, x = c - a (since x is a positive distance and c > a ) and

[tex] t = a - | X | = a - ( \frac{a^{2}}{c} ) = \frac{ac - a^{2}}{c} = \frac{a \cdot (c - a)}{c} .[/tex]

(Again, t is a positive distance and a > | X | . We are using the left branch of this horizontal hyperbola, which has complicated things as far as signs are concerned. For your presentation of the proof, you may want to use the right branch instead.)

We now see that

[tex] \frac{BH}{BG} = \frac{2a}{x} + 1 = \frac{2a}{c - a} + 1 = \frac{2a + c - a}{c - a} = \frac{c + a}{c - a} [/tex]

and

[tex] \frac{AH}{AG} = \frac{2a}{t} - 1 = \frac{2a}{\frac{a \cdot (c - a)}{c} } - 1 = \frac{2c}{(c - a) } - 1 = \frac{2c - c + a}{c - a} = \frac{c + a}{c - a} , [/tex]

thereby establishing the equality of the proportions. (Whoa!)

 

[Shackleford]

Ah! I completely forgot about those algebraic equations given in the textbook. I'm glad you caught the trick to make the Y/b² substitution into the hyperbola equation.

By the way, X = -a²/c. You forgot the negative when you distributed.

I worked it out all right. My algebra is a little different at the end, though. Thanks for the help!