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History of Math: Conics

  1. Sep 21, 2011 #1
    This is from my History of Mathematics course.

    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?
     
  2. jcsd
  3. Sep 22, 2011 #2
    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.
     
  4. Sep 22, 2011 #3

    dynamicsolo

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    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...
     
  5. Sep 23, 2011 #4
    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)
     
    Last edited: Sep 23, 2011
  6. Sep 23, 2011 #5

    dynamicsolo

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    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' .
     
  7. Sep 23, 2011 #6
    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).
     
    Last edited: Sep 23, 2011
  8. Sep 23, 2011 #7

    dynamicsolo

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    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!)
     
    Last edited: Sep 23, 2011
  9. Sep 24, 2011 #8
    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!
     
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