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 x-y-axes. Then the equation of the hyperbola is
| (1) |
x²
 a² |
| - | (y + k)²
 b² |
|
| = | 1 |
|
or,
| (1') |
b²x² - a²(y + k)² - a²b² = 0.
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Assume that the coordinates of the related points are as follows: M(0, 0), C(x1, y1), D(x2, y2), E(x3, y3), F(x4, y4), P(p, 0), Q(q, 0); and that the equations of lines DC and EF are y = m1x and y = m2x, respectively.
Substituting m1x into (1') gives
| (2) |
(b² - a²m1²) x² - 2a²m1kx - a² (b² + k²) = 0.
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The roots, x1 and x2, of equation (2) are the x-coordinates of C and D, where
| |
| x1 + x2 | = | | - | - 2m1ka²
 b² - m1²a² |
|
|
| and |
| x1 | | x2 | = | - a²(b² + k²)
 b² - m1²a² |
|
|
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Dividing these, we get
| |
m1x1x2
 x1 + x2 |
| = | - (b² + k²)
 2k |
|
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Similarly, by substituting y = m2x into (2) we will get
| |
m2x3x4
 x3 + x4 |
| = | - (b² + k²)
 2k |
|
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also (x3 and x4 are x-coordinates of E and F). Thus,
| (*) |
m1x1x2
 x1 + x2 |
| = | m2x3x4
 x3 + x4 |
|
|
which can further be rearranged (see note (I) below) to become
| (3) |
- x2x3
 m1x2 - m2x3 |
| = | x1x4
 m1x1 - m2x4 |
|
|
Observe that C, Q, F are collinear. The slope of QC and that of FQ are equal. So,
| |
y1
 x1 - q |
| = | - y4
 q - x4 |
|
|
Hence,
| (4) |
q - x1
 q - x4 |
| = | y1
 y4 |
|
| = | m1x1
 m2x4 |
|
|
q ≠ x1 and q ≠ x4.
Solving (4) for q, we have
| |
| q | = | (m1 - m2) x1x4
 m1x1 - m2x4 |
|
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In a similar manner, by equating the slopes of PE and DP we can get
| |
| p | = | (m1 - m2) x2x3
 m1x2 - m2x3 |
|
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Now by comparing p and q, and taking into account (3), it is easy to see that |p| = |q|. Therefore, MP = MQ.
Note
Derivation of (3): cross-multiplying and rearranging,
| | m1x1x2(x3 + x4) | = m2x3x4(x1 + x2) |
| | m1x1x2x3 - m2x3x2x4 | = m2x3x4x1 - m1x1x2x4 |
| | x2x3(m1x1 - m2x4) | = x1x4(m2x3 - m1x2), |
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.
Butterfly Theorem and Variants
- Butterfly theorem
- 2N-Wing Butterfly Theorem
- Better Butterfly Theorem
- The Lepidoptera of the Circles
- The Lepidoptera of the Quadrilateral
- The Lepidoptera of the Quadrilateral II
- Butterflies in Ellipse
- Butterflies in Hyperbola
- Butterflies in Quadrilaterals and Elsewhere
- Pinning Butterfly on Radical Axes
- Shearing Butterflies in Quadrilaterals
- The Plain Butterfly Theorem
- Two Butterflies Theorem
- Two Butterflies Theorem II
- Two Butterflies Theorem III
- Algebraic proof of the theorem of butterflies in quadrilaterals
- William Wallace's Proof of the Butterfly Theorem
- Butterfly theorem, a Projective Proof
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Copyright © 1996-2012 Alexander Bogomolny
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