Ask for high school math elective 4 4 after school exercises

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The distance between the two fixed points is 6, and the sum of the squares of the distance from the point m to these two fixed points is 26to find the trajectory equation for point m.

Ideas: (1) The linear Oa, ob equation and elliptic equation can be used to find the unary equation satisfying the coordinates of points A and B, and then the square of the horizontal and vertical coordinates of A and B can be obtained and substituted.

1|oa|2+1|ob|2. Simplify it

2) by s aob=12|oa||ob|，1|oa|2+1|ob|2=a2+b2a2b2, the minimum value can be found according to the mean inequality, and then according to s 2aob=

14|oa|2|ob|2. Put |ob|2 translates to |oa|2, and then according to the ellipse, |oa|range to find the maximum area

Solution: (1) Let the elliptic equation be x2a2+y2b2=1, and when the slope of the line oa exists and is not 0, let the equation be y=kx, a, b are two points on the ellipse, and oa ob the line ob equation is y=-1kx

Let a(x1,y1),b(x2,y2), and substitute y=kx into x2a2+y2b2=1 to get x12=a2b2b2+a2k2, y12=k2a2b2b2+a2k2

Substituting y=-1kx into x2a2+y2b2=1 gives x22=a2b2k2a2+b2k2, y22=a2b2a2+b2k2 1|oa|2+1|ob|2=1x12+y12+1x22+y22=1a2b2b2+a2k2+k2a2b2b2+a2k2+

1a2b2k2a2+b2k2+a2b2a2+b2k2=a2+b2a2b2

When one of the slopes of the line oa,ob does not exist, then the slope of the other line is 0, and 1|oa|2+1|ob|2=1a2+1b2=a2+b2a2b2

In summary, 1|oa|2+1|ob|2 is a fixed value.

2）s△aob=12|oa||ob|，∴s△2aob=14|oa|2|ob|2

From (1) to 1|oa|2+1|ob|2=a2+b2a2b2≥21|oa|21|ob|2=2|oa||ob|

s△aob=12|oa||ob|≥a2b2a2+b2，∴s△aobmin=a2b2a2+b2．

s△2aob=14|oa|2|ob|2=14|oa|2(1a2+b2a2b2-1|oa|2)

14(1a2+b2a2b2|oa|2-1|oa|4), with |oa|, the value of this function is increasing.

oa|≤a，∴s△2aob≤14(1a2+b2a2×b2×a2-1a4)=14a2b2

s△aobmax=ab2

In summary, S aobmin=a2b2a2+b2, and s aobmax=ab2

Many high school students in the study of mathematics, the most headache is the graph problem, because not only than junior high school mathematics more difficult space geometry problems, but also increased function subjects, both of which need to draw difficult graphics, so in high school mathematics, the combination of numbers and shapes is very critical, many function geometry problems, with the graph solution method can be quickly solved, without using a large number of drafts to calculate, I suggest that you can go to the Internet to buy a set of "function geometry special drawing ruler" It should be very helpful for your high school math learning, we used to order it collectively at school, and we couldn't buy it in the store, I hope my solution can help you.

The distance between the two fixed points is 6, and the sum of the squares of the distance from the point m to the two fixed points is 26, and the trajectory equation for the point m is found.

Solution: Take the straight line through the two fixed points F1 and F2 as the X axis, and take the perpendicular line of the line segment F1F2 as the Y axis to establish a Cartesian coordinate system, and the sum of squares of the distance from F1(-3,0), F2(3,0), M(X,Y) to these two fixed points is 26

x-3)^2+y^2]+[x+3)^2+y^2]=262x^2+2y^2+18=26

x 2 + y 2 = 4 is the trajectory equation for m.

The trajectory of m is a circle with the origin o(0,0) as the center and 2 as the radius.

Draw the pictures yourself, it shouldn't be difficult to look at these paintings, right?

Multiply both sides by , i.e. 2=2 cos 1cos 3-2 sin 1sin 3= cos 1- 3sin 1, i.e. x 2+y 2=x- 3y

In polar coordinates, 2=x 2+y 2, cos =x, sin =y, so that they can be converted to Cartesian coordinates.

Let AD be the height of the triangle ABC, take the line L where BC is located as the X axis, and the line where AD is located as the Y axis to establish a Cartesian coordinate system.

a(0,3)b(a,0)c(a+4,0), let p(x,y) be the outer center of the triangle abc, then pa=pb=pc, pa 2=pb 2=pc 2, so there is (x-a) 2+y 2=(x-a-4) 2+y 2=x 2+(y-3) 2

From the solution a=x-2, substituting into , and then from x2=6y-5 0, (y 5 6).

The trajectory of the outer center of the triangle ABC is the opening upward, the axis of symmetry is the y-axis, and the vertex is the parabola of (0,5 6).

x 2+y 2 i.e. the square of the distance from the point (x,y) to the origin is given by (x-2) 2+y 2=3 and the trajectory of the point (x,y) is a circle with (2,0) as the center radius 3.

The closest point to the origin of this circle is 2-3 and the farthest point is 2+ 3, so the value range of x 2+y 2 is (7-4 3, 7+4 3) or x= 2 3-y 2) directly from (x-2) 2+y 2=3).

Generations x 2+y 2 get the function about y.

The value range of x 2+y 2 can be obtained from the value range of y, and the result is the same as above.

(x-2) 2+y 2=3 is a circle with (2,0) as the center of the circle and 3 as the radius, if the center of the circle is translated to (0,0), then the circle equation is x 2+y 2=3, so x 2+y 2=3

Khan, this is the original question --- textbook

After taking ---1 DE is the median line, df BG=AF AG=FE GC deformation, BG GC=DF FE

2 Look at the similarity of GOB and FEO, and the similarity of GOC and DEO, that is, BG GC=EF DF=DF FE (with the first step), so BG=GC