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.
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