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A math problem in science in senior three.
Solution: (i) F2 (1, 0) About the line L: X-2Y+4 = 0 Symmetric point G (- 1, 4)

The intersection p of GF 1 and l is on an ellipse,

∴2a=|pf 1|+|pf2|=|gf 1|=4.

∴b2=a2-c2=3.

So the elliptic equation is X24+Y23 = 1.

(II) It is known from the conditions that the slopes of the straight lines PM and PN exist and are not 0,

Easy to get point p (- 1, 3 2), let the equation of straight line PM be y = k (x+1)+3 2,

Y is eliminated simultaneously by elliptic equation and linear PM equation,

Arrange (4k2+3) x2+4k (2k+3) x+4k2+12k-3 = 0,

∵P On the ellipse, the two roots of the equation of∴ are 1 and x 1.

∴ 1? x 1=? 4k2+ 12k? 3 4k2+3,x 1=? 4k2+ 12k? 3 4k2+3,

The inclination angles of the lines PM and PN are complementary,

The slopes of the straight lines PM and PN are opposite to each other.

∴x2=? 4k2? 12k? 3 4k2+3。

So x 1? x2=? 24k 4k2+3,x 1+x2=6? 8k2 4k2+3。

And y1= k (x1+1)+32, y2 =? k(x2+ 1)+3 2,

∴y 1-y2=k(x 1+x2+2)= 12k 4k 2+3。

The slope of a straight line MN kmn = y 1? y2 x 1? x2 =? 12 (fixed value)