Math problems in the sophomore year of high school, math problems in the sophomore year of high scho

By the question, there is |f(-1)|= |-a+b|<=1 ; f(1)|= |a+b| <=1

1<= -a+b<=1 ;1<= a+b<=1 The addition of the two formulas has 1<= b<=1 , i.e.: |b|1 is added by the first equation (multiplication -1) with -1< = a-b<=1 and the second equation. There is -1< = a<=1, i.e. |a|≤1

Supplement: Yashemao is not wrong, the selection of linear problems is very particular, and it is generally necessary to consider the situation at both ends of a given interval, such as this problem, with |f(-1)|= |-a+b|<=1 ; f(1)|= |a+b|<=1, you can guarantee "dang|."x|At 1 o'clock, there are |f(x)|1", that is, the two inequalities used here are equivalent to the conditions of the title. What you say, "There must be such a line segment:."

1<=a<=1,-1<=b<=1", not to use the |s at both endsf(-1)|= |-a+b|<=1 ; f(1)|= |a+b|<=1 solution. So I don't think there's a problem with such a solution, if you use |f(0)|<=0 is problematic to calculate the range of b.

Let the equation for the circle be (x-2) 2+(y-y0) 2=m and substitute (-1,5),(6,-2) to get the system of equations.

9+(5-y0)^2=m

16+(-2-y0)^2=m

The top and bottom phases are subtracted.

y0=1 to get m=25

So (x-2) 2+(y-1) 2=25

1 (1) The length of the major axis is 4, i.e., 2a=4, and the eccentricity is 2/2 of the root number 3 is equal to e=c a, so c = root number 3, a 2 = b 2 + c 2, b = 1, so the equation is x 2 4 + y 2 = 1

2) The line segment ab is a string of the ellipse e, and the straight line l is perpendicular to the chord ab, we can know that the slope of the straight line of ab is -1, and the point a(0,1), we can write the equation of the line ab and the equation of the ellipse to solve the coordinates of the point b. The bisector of the line l shows that the midpoint of ab is on the line l, and m can be solved by substituting the coordinates of the midpoint of ab into the equation of l.

2 to the absolute value of -22, then 2x0 2+0m-1<0, constant less than zero. Therefore, the range of values of m is m=.

, the length of the long axis is 4, and the eccentricity is the root number of 2 points c a = the root number of 2 points 3 c = the root number 3 b = 1

1.The standard equation for the ellipse e x 2 4 + y 2 = 1

2.Straight line l perpendicular bisector chord ab kab=-1 b(x0,y0) a(0,1) y0-1 x0=-1 x0=1-y0 substitution into the ellipse.

x0=7 4, y0=-3 4 ab midpoint is on l:y=x+m, m=-3 2

1) a: -1 (2) c = a cross b 0< = x<2 2x 2 + mx-1<0 x = 0 set up x=2 7 + 2 m<0 m<-7 2

1.(1)a=2;c = root number 3, b = 1

Equation: x 2 4 + y 2 = 1

2) The slope of ab is -1, y=-x+1, and if b(x,-x+1), then the midpoint of ab (x 2, -x 2+1) is on l.

x 2 + 1 = x 2 + m gives x + m = 1

and b on the ellipse x 2 4 + (1-x) 2 = 1, x = 0 or x = 8 5m = -3 , -3 5).

1.When the p point is at the upper or lower vertex of the ellipse, the cosine value is the smallest, and according to the cosine theorem, the lowest value is -(5 13).

Anyway, my homework at home happened to be the same problem...

Let the equation for the ellipse be x 2 a 2 + y 2 b 2 = 1, then a = 6 2 = 3 b 2 = a 2-c 2 = 9-8 = 1

The equation for an ellipse is x 2 9 + y 2 = 1

If the ab ab abscissa is x1, x2, then the ab coordinates are a(x1,x1+2), b(x2,x2+2).

The coordinates of the midpoint of ab are ((x1+x2) 2,(x1+x2) 2+2), and y=x+2 is substituted into the elliptic equation.

10x^2+36x+27=0

Since ab is the intersection of a straight line and an ellipse, the two roots of the above equation are ab's abscissa and are known by Vedder's theorem.

x1+x2=-36/10=-18/5

Therefore, the coordinates of the midpoint of AB are (-9 5, 1 5).

Children, it's not our duty to help you do the problems, hehe, practice more by yourself Mathematics is practiced.

First, write the elliptic equation from the focal coordinates and the major axis, and then calculate the coordinates of the intersection points a and b together with the linear equations, and the midpoint can be calculated.

Contemptible people have always given him fishing, and I hope you will make your own calculations and deepen your impression.

The ellipse x a +y b =1(a>b>0), f1, f2 are the left and right focal points of the ellipse, if f2 is the vertex, the parabola with f1 as the focal point passes the upper and lower vertices of the ellipse, and the value of b a is obtained.

Ellipse x a + y b = 1 (a>b>0) --c = a -b .1)

> parabolic vertex f2(c,0), p 2=2c--- parabolic equation: y =-2p(x-c)=8c(c-x).

0, b) on the parabola--- b = 8c (c-0) = 8c --c = b 8....2)

1) (2) Synthesis: b 8 = a -b --8a = 9b --b ; a^=8:9---b/a=2√2/3