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Conic Triangle (Posted on 2014-10-17) Difficulty: 3 of 5

  
A line tangent to a hyperbola intersects its asymptotes at points L and N.
Let M be the intersection of the asymptotes.

Prove that the area of ΔLMN is a constant.
  

  Submitted by Bractals    
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Solution: (Hide)

  
The central equation for a hyperbola is

    x2     y2
   ---- - ---- = 1                                              (1)
    a2     b2
where 2a is the distance between the vertices and

   y = ±bx/a                                                    (2)
are the equations of the asymptotes.

Differentiating equation (1) gives

         b2x
   y' = -----                                                   (3)                                                          
         a2y
Because of the symmetry of the hyperbola, we only
need to consider the point of tangency (xt,yt) in
the first quadrant. The tangent line

   y = y'tx + k

        b2xt
     = ----- x + k                                                                                                               
        a2yt

         b2xt2 
   yt = ------- + k                                                                                                               
         a2yt

        b2xt           b2xt2
   y = ----- x + yt - ------                                                                                                          
        a2yt           a2yt

   a2yty = b2xtx + a2yt2 - b2xt2

   a2yty = b2xtx - a2b2

        b2xt      b2
   y = ----- x - ---                                            (4)                                                                                                          
        a2yt      yt
Equation (4) and y = bx/a intersect at point

   L = (xl,yl) = ( a2b/[bxt - ayt] , ab2/[bxt - ayt] )            (5)
which lies above the x-axis.

Equation (4) and y = -bx/a intersect at point

   N = (xn,yn) = ( a2b/[bxt + ayt] , -ab2/[bxt + ayt] )           (6)
which lies below the x-axis.

Equation (4) and x-axis intersect at point

   I = (xi,yi) = ( a2/xt,0 )                                     (7)
We can now calculate the area.

   Area(ΔLMN) = Area(ΔLMI) + Area(ΔNMI)

              = (a2/2xt)(ab2/[bxt - ayt] ) + (a2/2xt)(ab2/[bxt + ayt] )  

              = (a3b2/2xt)([bxt + ayt + bxt - ayt] /[b2xt2 - a2yt2 ])

              = (a3b2/2xt)(2bxt/a2b2)

              = ab                                              (8)
Note:

The above argument does not hold if the point of tangency is the
vertex (a,0) because of equation (3). But, the area for this case is
easily calculated.

   Area(ΔLMN) = (1/2)(base)(altitude)

              = (1/2)|LN||MI|

              = ab                                              (9)
Equations (8) and (9) agree on the constant ab.

QED

Note:

See Ady's post for how simple the problem is for a rectangular
hyperbola.
  

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Stretching a point?Bractals2014-10-20 11:25:36
Some ThoughtsStretching a point?Harry2014-10-19 19:01:45
SolutionSOLUTIONAdy TZIDON2014-10-18 04:16:30
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