Discover several interesting math problems about the equations of straight lines and circles in high school courses.

the butterfly theorem

Theorem content: The midpoint M of the chord PQ on the circle O, the intersection point M is defined as two chords AB and CD, and the chords AD and BC intersect PQ on X and Y respectively, then M is the midpoint of XY.

It is proved that O passing through the center of the circle is the vertical line between AD and BC, and the vertical foot is S and T, connecting OX, OY, OM, SM and MT.

∫△AMD∽△CMB

∴AM/CM=AD/BC

∫SD = 1/2AD，BT= 1/2BC

∴AM/CM=AS/CT

∠∠A =∠C

∴△AMS∽△CMT

∴∠MSX=∠MTY

∠∠OMX =∠OSX = 90 degrees

∴∠OMX+∠OSX= 180

∴O, s, x, m four-point * * * circle

Similarly, O, T, Y and M are four-point * * * circles.

∴∠MTY=∠MOY,∠MSX=∠MOX

∴∠MOX=∠MOY，

∵OM⊥PQ

∴XM=YM

This theorem also holds in ellipses, as shown in the figure.

1, the major axes A 1 and A2 of the ellipse are parallel to the X axis, and the minor axis B 1B2 is on the Y axis with the center of M(o, r) (b >; r & gt0)。 (1) Write the equation of ellipse, and find the focal coordinates and eccentricity of ellipse.

(ii) The straight line y=k 1x intersects the ellipse at two points C(x 1, y 1) and D(x2, y2) (y2 >; 0); The straight line y=k2x intersects the ellipse at two points G(x3, y3), H(x4, y4) (y4 >; 0)。 Verification: k1x1x 2/(x1+x2) = k2x3x4/(x3+x4)

(Ⅲ) For C, D, G and H in (Ⅱ), let CH intersect with X axis at point P and GD intersect with X axis at point Q ... Prove: | OP | = | OQ |.

(The proof process does not consider the case that CH or GD is perpendicular to the X axis. ) The reference answers given by experts from the Admissions Examination Office of Beijing Education Examinations Institute in the Collection of Answers to the 2003 National Unified Entrance Examination for Colleges and Universities are as follows:

(18) This topic mainly examines the basic knowledge of straight lines and ellipses, and examines the ability to analyze and solve problems. 15. (1) solution: the elliptic equation is x2/a2+(y-r)2/b2= 1, and the focal coordinate is

(2) prove the equation y=k of straight line CD? Substituting x into the elliptic equation gives b2x2+a2(k 1x-r)2=a2b2.

Finishing, (b2+A2k12) x2-2k12rx+(a2r2-a2b2) = 0.

According to Vieta's theorem, you must

x 1+x2 = 2k 1a2r/(B2+a2k 12)，x 1 x2 =(a2r 2-a2 B2)/(B2+a2k 12)，

So x1x2/(x1+x2) = (R2-B2)/2k1R1.

Substituting the equation y=k2x of the straight line GH into the elliptic equation, the same can be obtained.

x3x4/(x3+x4)=( r2-b2)/2k2r ②

K1x1x2/(x1+x2) = (R2-B2/2r = K2X3x4/(x3+x4) from ① and ②.

So the conclusion is established.

(3) Proof: Set point P(p, o) and point Q(q, o).

From the lines of c, p and H * * * *, (x1-p)/(x4-p) = k1x1/k2x4.

The solution is p = (k1-k2) x1x4/(k1x1-k2x4).