Mathematics Elective 1 1, Mathematics Compulsory 2, all formulas

It's better to summarize it yourself, so that you can understand it more deeply!

x-a)^2+(y-b)^2=r^2

Elliptical fissure: x 2 a 2 + y 2 b 2 = 1 (x-axis) (a>b>0).

x 2 b 2+y 2 a 2 = 1 (y-axis) no shirt (a>b>0) e = c a (0

x-a)^2+(y-b)^2=r^2

Ellipse: x 2 a 2 + y 2 b 2 = 1 (x-axis) (a>b>0) x 2 b 2+y 2 a 2=1 (y-axis) (a>b>0) e=c a(00).

The line that crosses the focus intersects the parabola at two points:

x1*x2=p^2/4 y1*y2=-p^2

It's useless to memorize formulas, so it's better to read books and do real questions.

（x-a)^2+(y-b)^2=r^2

Ellipse: x 2 a 2 + y 2 b 2 = 1 (x-axis) (a>b>0) x 2 b 2+y 2 a 2=1 (y-axis) (a>b>0) e=c a(00).

The line that crosses the focus intersects the parabola at two points:

x1*x2=p^2/4y1*y2=-p^2

y=2x^3-3x^2-12x+5

y'=6x 2-6x-12=6(x+3)(x-4)y has extremums at x=-3 and x=4.

Then the extreme values at [0,2] are x=0 and x=2

y(0)=5, y(2)=16-24-24+5=-27, so the maximum and minimum values of y=2x 3-3x 2-12x+5 on [0,2] are 5 and -27, respectively

Solution: y=2x 3-3x 2-12x+5

y'=6x^2-6x-12

On [0,2] on y'is negative, so the function is decreasing, i.e. y=5 when the maximum value is x=0, and y=-15 when the minimum value is x=2

If the point p is on the ellipse 7x +4y=28, then the maximum distance from the point p to the straight line 3x-2y-16=0 is .

Solution: The slope of the straight line 3x-2y-16=0 k=3 2

Derivative of the elliptic equation: 14x+8yy =0, so y =-14x 8y=-7x 4y

Let -7x 4y=3 2, and get y=-7x 6

Substituting the elliptic equation: 7x +4(-7x 6) = 7x +(49 9)x =112x 9=28

112x =252, x =252 112=9 4, so x= 3 2Correspondingly, y= (7 40).

That is, the tangents of the dots (3 2, -7 4) and (-3 2, 7 4) on the ellipse are parallel to the straight line, and the image of the ellipse and the straight line.

It can be seen that the point (-3 2, 7 4) should be taken as the p point, and the distance d from this point to the straight line is the maximum.

dmax = 3 (-3 2)-2 (7 4)-16 13=24 13=(24 13) 13 can be used as graph verification).

The equation for the ellipse is miswritten.

1. The first point (1 3, 1 9) is on the curve y=f(x), so there is: 1 9=a 27+b 9

F. again'(x)=3ax 2+2bx and parallel to the x-axis, i.e., 0=a 3+2b 3

The solution yields a=-6 and b=3

2. y=f(x) extremum, i.e., f'(x)=0. Replace a = -6 and b = 3 with f'(x)=3ax 2+2bx=0, and the solution is x=1 3 and x=0. When x=1 3, y=1 9;When x=0, y=0.

So the minimum value of the function is 0.

Indicates that the origin is on a circle, and the distance between the center of the circle (a, b) and the origin (0,0) is equal to the radius r. The formula for the distance between the two points is as follows: under the root number (a + b) = r. That is, the sufficient condition is a +b =r

1 All the answers are b. Since the flagpole is perpendicular to the ground, it may be advisable to set it to PO, so observe points a, b, and c 、...... on the ground where the elevation angle of the top p is equalIt forms a triangle with PO, 、...... APO, BPO, and CPOThese triangles are congruent, so ao bo co ......i.e. points a, b, c, 、......The distance to the point o is equal, so the points a, b, c 、......On a circle with o as the center.

b The distance to the fixed point is equal to the fixed length.