Butterflies in Hyperbola
Sidney H. Kung
We give an analytic proof of the Butterfly Theorem for hyperbolas.
Let M(0, k) be the midpoint of a chord AB parallel to the major axis of a hyperbola. Through M two other chords CD and EF are drawn. ED cuts AB at P and CF cuts AB at Q. Then M is also the midpoint of PQ.
Through M, we introduce a new set of x-y-axes. Then the equation of the hyperbola is
|(1')||b²x² - a²(y + k)² - a²b² = 0.|
Assume that the coordinates of the related points are as follows:
Substituting m1x into (1') gives
|(2)||(b² - a²m1²) x² - 2a²m1kx - a² (b² + k²) = 0.|
The roots, x1 and x2, of equation (2) are the x-coordinates of C and D, where
|x1 + x2||=|
Dividing these, we get
also (x3 and x4 are x-coordinates of E and F). Thus,
which can further be rearranged (see note (I) below) to become
Observe that C, Q, F are collinear. The slope of QC and that of FQ are equal. So,
q ≠ x1 and q ≠ x4.
Solving (4) for q, we have
In a similar manner, by equating the slopes of PE and DP we can get
Now by comparing p and q, and taking into account (3), it is easy to see that
Derivation of (3): cross-multiplying and rearranging,
m1x1x2(x3 + x4) = m2x3x4(x1 + x2) m1x1x2x3 - m2x3x2x4 = m2x3x4x1 - m1x1x2x4 x2x3(m1x1 - m2x4) = x1x4(m2x3 - m1x2),
The above results are valid for parabolas. To verify, one may use the equation
x² = p(y + k), p, k > 0,to start.
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