Butterflies in Hyperbola
Sidney H. Kung
We give an analytic proof of the Butterfly Theorem for hyperbolas.
Theorem
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.
Proof
Through M, we introduce a new set of xyaxes. Then the equation of the hyperbola is
(1) 

or,
(1')  b²x²  a²(y + k)²  a²b² = 0. 
Assume that the coordinates of the related points are as follows:
Substituting m_{1}x into (1') gives
(2)  (b²  a²m_{1}²) x²  2a²m_{1}kx  a² (b² + k²) = 0. 
The roots, x_{1} and x_{2}, of equation (2) are the xcoordinates of C and D, where
x_{1} + x_{2}  = 

x_{1} 

Dividing these, we get
 = 

Similarly, by substituting y = m_{2}x into (2) we will get
 = 

also (x_{3} and x_{4} are xcoordinates of E and F). Thus,
(*) 

which can further be rearranged (see note (I) below) to become
(3) 

Observe that C, Q, F are collinear. The slope of QC and that of FQ are equal. So,
 = 

Hence,
(4) 

q ≠ x_{1} and q ≠ x_{4}.
Solving (4) for q, we have
q  = 

In a similar manner, by equating the slopes of PE and DP we can get
p  = 

Now by comparing p and q, and taking into account (3), it is easy to see that
Note
Derivation of (3): crossmultiplying and rearranging,
m_{1}x_{1}x_{2}(x_{3} + x_{4}) = m_{2}x_{3}x_{4}(x_{1} + x_{2}) m_{1}x_{1}x_{2}x_{3}  m_{2}x_{3}x_{2}x_{4} = m_{2}x_{3}x_{4}x_{1}  m_{1}x_{1}x_{2}x_{4} x_{2}x_{3}(m_{1}x_{1}  m_{2}x_{4}) = x_{1}x_{4}(m_{2}x_{3}  m_{1}x_{2}), implying (3).
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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