It is known that the proportional function y kx intersects the parabola y ax2 3 at 2,4 .

The proportional function y=kx intersects with the parabola y=ax 2+3 at (2,4)=》2k=4 4a+3=4

k=2, a=1/4

1) Proportional function analytical: y=2x

and parabolic analytical: y=1 4x 2+3

2) Point A is (1 4, 1 2).

Point b is (0,3).

Origin o(0,0).

The distance from point A to Bo is 1 4, and the length of Bo = 3 = Area = 1 2 * 1 4 * 3 = 3 8

1. Substituting the point (2,4) into y=kx and y=ax 3 obtains: k=2, a=(1 4), then: y=2x, y=(1 4)x 3

2. The point on the image of the proportional function is a(1 4,1 2), the parabolic vertex is b(0,3), and the origin is o(0,0), then the area of the triangle AOB is s=(1 2) bo h=(1 2) 3 (1 4)=3 8

We know that the analytic formula of the proportional function y=kx, so k=4 2=2, so the analytic formula of the proportional function y=2x, and then substitute (2,4) into y=ax 3, we get a=1 4, so the analytic formula of the parabola is y=(1 4)x 3;

Since the ordinate of point A on the image of the proportional function is 1 2, so 1 2 = 2x, and the solution is x=1 4, so the coordinates of point A are (1 4, 1 2), and the vertex coordinates are obviously (0,3) origin (0,0), and since the distance from point A to Bo is 1 4, and the length of Bo is 3, the area of the triangle AOB is s=(1 2)x3x(1 4)=3 8

Solve the equation and guess the bad limb.

y=x 2-2x+4,y=kx,x,x2-2x+4=kxx 2+(-2-k)x+4=0, discriminant = 2-4ac(-2-k) 2-4*4

k 2 + 4k - 12 > Li Zhaoming 0

So k2

k=4 gives the simultaneous equation of y=4x 2 and y=kx-1 of Yumu:

y=4x^2（1）

y=kx-1（2）

Substitute (2) with (1).

4x^2-kx+1=0

Again, the parabolic key y=4x 2 has a unique intersection point with the straight line y=kx-1, that is, the equation has a unique manuscript solution.

then, the formula gets k = 4

This question seems complicated, but it is actually very simple, and if you think about it carefully, you will find that the parabola mentioned in the known conditions is simply confusing: because OA is determined by y=kx, although the point p is on the parabola, it can be regarded as a recognition point from the intersection of OA and X axis in the back; The next step is to pass Q as the PM perpendicular, which has nothing to do with the parabola, so you can directly ignore the parabola when solving the problem!

The specific proof is as follows:

The X axis and the Y axis are perpendicular lines respectively through Q, and the X axis is at E and the Y axis is at F;

No is perpendicular to OM and NQ is perpendicular to MQ, so NOMQ is a circle of four points; So the angle qnf = the angle qme;

and the angle qfn=angle qem=90 degrees, so the triangle qnf is similar to the triangle qme, so qm qn=qe qf;

Q is on y=kx, so qm qn=qe qf=y x=k;

It is also known that a(3,6) is on y=kx, and substituting 6=3k gives k=2;

Therefore, qm qn=k=2;

(1) Substituting the point a into the linear equation to get k=2 y=2xoa=3, root number 5

2) The point Q is the perpendicular line of the x-axis and Y-axis, and the vertical foot is g and h

qnh∽△qmg qm/qn=qg/qh=qg/og=k=2(3)∠bae=∠bed=∠aod

Description ABE OED

bae= aod indicates that the point b is fixed first find the point b and set b(x1,y1), extend the x-axis of ab to the point q(x2,0), aod= tan =2

The slope angle of the straight line aq is 2, and the solution of x2=15 2 is the linear aq equation y=-4 3(x-15 2)=-4 3x+10

The intersection points of the straight line aq and the parabola are a and b, and the simultaneous equation is x1=6 and y1=2

y=kx-2 y=kx-4k+4 y=4x y=4x δ=4k+4) refers to -16k 0 k≠ with 0 k 1 2 k≠0 k 1 2 and k ≠ 0

y=2x2x2+4x+k-1=2(x+1) 2+k-3Because the parabola opens upward, the condition that there are two intersections with the x-axis is that the minimum value of y is less than 0, so y=k-3<0

i.e. k<3

y=2x +kx-4 and the intersection points with the x-axis are x1 and x2, so x1 and x2 are the roots of 2x +kx-4=0.

So x1+x2= -k 2,x1x2= -2 because Wu Zhijian 1 x1+1 x2=(x1+x2) (x1x2)=-3 2, so k 4=-3 2,k=-6

Because the abscissa of point d = (x1+x2) 2=-k 4=3 2, d(3 2,-17 2).

Point C coordinates (0,-4).

Do c'(0,4) and connect c'd, which intersects with the x-axis at e, where the value of CE+DE is the smallest.

Proof: If e is not at this point, because c'e=CE, so the ECD will form a triangle, and the sum of the two sides of the triangle is greater than the third side.

The expression for c'd: y=(-3 2) (4+17 2) x+4=-3 25 x+4

When y=0, x=100 3

So m=100 3