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Pudong New Area 20 15 Mathematical Module 2
( 1)∵C(0,-3),

∴OC=3.y= 14x2+bx-3.

OA = 2OC,

∴OA=6.

∫A = 14 > 0, point a is on the right side of point b, and the intersection of parabola and y axis is C(0, -3).

∴A(6,0).

∴0= 14×36+6b-3,

∴b=- 1.

∴y= 14x2-x-3,

∴y= 14(x-2)2-4,

∴M(2,-4).

A: The analytical formula of parabola is y= 14x2-x-3, and the coordinate of m is (2,-4);

(2) As shown in figure 1, the intersection point M is MH⊥x axis, the vertical foot is H point, the intersection point AC is N point, the intersection point N is NE⊥AM is E point, and the vertical foot is E point.

∴∠AHM=∠NEM=90。

In Rt△AHM, HM=AH=4, which is obtained from Pythagorean theorem.

AM=42,

∴∠AMH=∠HAM=45。

Let the analytical formula of straight line AC be y=kx+b, as can be seen from the meaning of the question

0=6k+b? 3=b,

Solution: k = 12b =? 3,

The expression of straight line AC is y = 12x-3.

When x=2 and y=-2,

∴N(2,-2).

∴MN=2.

∠∠NEM = 90 degrees, ∠NME=45 degrees,

∴∠MNE=∠NME=45,

∴NE=ME.

In Rt△MNE,

∴NE2+ME2=NM2,

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