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If the two focuses and the two vertices of the minor axis of the ellipse are the four vertices of a diamond with a 60 ° angle, then the eccentricity of the ellipse is 0 A. 1 / 2 b. √ 3 / 2 C. √ 3 / 3 d.1/2 or √ 3 / 2
Let the focus be on the x-axis, and let the intersection of the ellipse and the y-axis M1, M2, then
The angle m1f1m2 is 60 ° 2A = 2B + 2B, C = root 3B
Then E = C / a = radical 3B / 2B = radical 3 / 2
If the focus is on the y-axis, it is still 3 / 2 of the root
So choose B
Let p be the point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1. If F1 and F2 are the two focuses of the ellipse, then the absolute value Pf1 + absolute value
For the ellipse X & # 178 / 25 + Y & # 178 / 16 = 1, where a = 5, B = 4
|PF1|＋|PF2|=2a=10
Given that the four vertices of square ABCD are on the ellipse X / (a ^ 2) + Y / (b ^ 2) = 1 (a > b > 0), AB is parallel to the X axis, and ad passes through the left focus F, then the ellipse
What is the centrifugal rate?
Because AB is parallel to the X axis, the side length of the square is equal to the focal length 2C
When ad passes the left focus F1, the length of AF1 is equal to half of the side length, | AF1 | = C
Let the right focus of the ellipse be F2,
According to Pythagorean theorem | af2 | = √ (C & sup2; + 4C & sup2;) = √ 5C
According to the definition of ellipse: | AF1 | + | af2 | = 2A
That is, C + √ 5C = 2A, C / a = √ (5-1) / 2
The centrifugal rate is √ (5-1) / 2
The focus F1, F2 and point P of the ellipse x ^ 2 / 9 + y ^ 2 / 2 = 1 are on the ellipse. If the absolute value of Pf1 = 2 ~ then the absolute value of PF2 = angle f1pf2, the size of PF2 is
RT
a²=9
A=3
Defined by ellipse
|PF2|+|PF1|=2a=6
|PF2|=4
b²=2
c²=9-2=7
c=√7
F1F2=2c=2√7
From cosine theorem
cosF1PF2=(PF1²+PF2²-F1F2)/(2PF1*PF2)=-1/2
So the angle is 2 π / 3
If the square ABCD is known, then the eccentricity of the ellipse passing through C and D is 0
The line passing through a and B is the x-axis, the vertical bisector of AB is the y-axis, and the midpoint of AB is o
Draw a graph, AC + CB = 2A, ab = 2c, let the square side length be 1, then AB = 1, AC = root 2, so the eccentricity is equal to a / C = (1 + root 2) / 1, equal to 1 + root 2
Let AB be the chord passing through the center of the ellipse x ^ 2 / 9 + y ^ 2 / 25 = 1, and F1 be the focus on the ellipse. The parametric equation is used to find the maximum area of △ Abf1
a=3,b=5
s(max)=(1/2)*2*3*5=15
If the ellipse with the relative vertex A and C of the square ABCD as the focus just passes through the midpoint of the four sides of the square, the eccentricity of the ellipse is ()
A. 10−23B. 5−13C. 5−12D. 10−22
Let the side length of the square be 2 and the center of the square be the origin, then the elliptic equation is x2a2 + y2b2 & nbsp; = 1 and C = 2 A2-B2 = C2 = 2. ① the coordinates of the midpoint of the BC side of the square are (12, 12). Substituting the equation into the equation, we get 12a2 + 12b2 = 1. ② the simultaneous solution of the equation is a = 1 + 52 e = CA = 10-22
Let a be the moving point chord AB and AC on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, passing through the focus F1 and F2 respectively. When AC is perpendicular to the X axis, there is exactly one
Let a be a moving point on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, and the chords AB and AC pass through the focus F1 and F2 respectively. When AC is perpendicular to the X axis, there is exactly | AF1 |: | af2 | = 3:1, (1) find the eccentricity of the ellipse
(2) Let AF1 = mf1b, af2 = nf2c, and prove that M + n is the fixed value 6
Let a be a moving point on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, and the chords AB and AC pass through the focus F 1 and F 2 respectively. When AC is perpendicular to the X axis, there is exactly | AF1 |: | af2 | = 3:1
Find: (1) find the eccentricity of ellipse (2) let vector AF1 = m, vector F1B, vector af2 = n, vector F2C, prove that M + n is the fixed value 6
(1) Let AF1 = 3x, then af2 = X
Then 3x + x = 2A, x = A / 2
So AF1 = 3A / 2, af2 = A / 2
According to Pythagorean theorem
(3a/2)²=(a/2)²+(2c)²
4c²=2a²
c²/a²=1/2
e²=1/2
So eccentricity e = √ 2 / 2
(2) C & sup2; = 1 / 2A & sup2;, a & sup2; = B & sup2; + C & sup2;, so B = C
The elliptic equation becomes: X & sup2; + 2Y & sup2; = 2B & sup2;
Focus coordinates F1 (- B, 0), F2 (B, 0)
Set point a (x0, Y0) B (x1, Y1) C (X2, Y2)
M = vector AF1 / vector F1B, n = vector af2 / vector F2C
M = - Y0 / Y1, n = - Y0 / Y2
AC slope of straight line: Y0 / (x0-b)
Let the linear AC equation: x = [(x0-b) / Y0] y + B
Substituting into the elliptic equation: X & sup2; + 2Y & sup2; = 2B & sup2;
Sorting: (3b-2x0) y & sup2; + 2 (x0-b) y0y-by0 & sup2; = 0 (Note: x0 & sup2; + 2y0 & sup2; = 2B & sup2;)
Weida theorem: y0y2 = - by0 & sup2; / (3b-2x0)
y2=-by0/(3b-2x0)
n=-y0/y2=(3b-2x0)/b
The equation of line AB: x = [(x0 + b) / Y0] y-b
Substituting into the elliptic equation: X & sup2; + 2Y & sup2; = 2B & sup2;
Sorting: (3b + 2x0) y & sup2; - 2b (x0 + b) y-by0 & sup2; = 0
Weida's theorem: y0y1 = - by0 & sup2; / (3b + 2x0)
y1=-by0/(3b+2x0)
m=-y0/y1=(3b+2x0)/b
m+n=(3b+2x0)/b+(3b-2x0)/b=6b/b=6
When the AC slope does not exist, the AC is perpendicular to the x-axis
Y0 = - Y2, so n = 1
x0=b,m=(3b+2b)/b=5
M + n = 6 also holds
The proof is complete
In the plane rectangular coordinate system, the vertex o of the rectangular oacb is at the origin of the coordinate, the vertices a and B are on the positive half axis of the x-axis and y-axis respectively, OA = 3, OB = 4, and D is the midpoint of the edge ob
(1) Find the length of line CD;
(2) If e is a moving point on the edge OA, the minimum perimeter of △ CDE is obtained;
(3) If e and F are two moving points on the line OA (point E is to the left of point F), and EF = 2, when the perimeter of quadrilateral cdef is the smallest, the coordinates of points E and F are obtained
3. Take cm = 2 on CB, then M (1,4), take symmetric point n (0, - 2) of D on the negative half axis of Y axis, connect Mn with X axis at point E, connect de and CF, then the perimeter of quadrilateral cdef is the smallest. (draw the picture according to my description) because CD and EF are fixed values, as long as de + CF is the smallest
If AB passes through the chord at the center of the ellipse & nbsp; X225 + y216 = 1 and F1 is the focus of the ellipse, then the maximum area of △ f1ab is ()
A. 6B. 12C. 24D. 48
Let a's coordinates (x, y) be: B (- x, - y), then △ f1ab area s = 12of × | 2Y | = C | y |. When | y | is the largest, △ f1ab area is the largest. From the graph, when point a is at the vertex of the ellipse, its △ f1ab area is the largest, then the maximum value of △ f1ab area is: CB = 25 − 16 × 4 = 12
RELATED INFORMATIONS
- 1. In the plane rectangular coordinate system, the vertex o of the rectangular oacb is at the origin of the coordinate, the vertices a and B are respectively on the positive half axis of the x-axis, OA = 3, OB = 4, and D is the center of the edge ob In the plane rectangular coordinate system, the vertex o of the rectangular oacb is at the origin of the coordinate, the vertices a and B are respectively on the positive half axis of the X axis, OA = 3, OB = 4, and D is the midpoint of the edge ob. If e and F are two moving points on the edge OA and EF = 2, then the minimum perimeter of the quadrilateral cdef is the coordinates of points E and F
- 2. As shown in the figure, in the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length of 1 are on the Y-axis and X-axis respectively, As shown in the figure, in the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length of 1 are respectively on the positive half axis of Y axis and X axis, and point O is at the origin. Rotate the square oabc clockwise around point O, and stop rotating when point a falls on the line y = x for the first time. In the process of rotation, intersect the straight line y = x at point m, and BC intersect the X axis at point n ）In the process of rotation, when Mn and AC are parallel, calculate the rotation degree of oabc ）&If the circumference of △ BMN is p, does the value of P change during the rotation of oabc? Please prove your conclusion; ）Let Mn = m, what is the minimum area of △ mon when m is, and what is the minimum value? Then the radius of the inscribed circle of △ BMN is obtained
- 3. If the image of the linear function y = (2-m) x + m passes through the first, second and fourth quadrants, then the value range of M is______ .
- 4. The first-order function y = (2m-1) + X + (3 + n) is used to find the value range of m.n through one two three quadrants
- 5. When the value of the first-order function y = (4 + 2m) x + M-4 is known, the image passes through one three four quadrants
- 6. Find the coordinates of the intersection point of the image of the function y = 2x + 1 and y = - 3x-4
- 7. If the point (m, n) is on the graph of function y = 2x + 1, then the value of 2m-n is () A. 2B. -2C. 1D. -1
- 8. The eccentricity of the ellipse with focus on the x-axis is two-thirds root sign three, and the ellipse intersects with (X-2) ^ + (Y-1) ^ = 5 / 2 at two points a and B. the length of line AB is equal to the diameter of the circle 1. Find the equation of line ab 2. Find the equation of ellipse
- 9. Given the proposition p: "equation x2 + y2m = 1 represents the ellipse with focus on Y axis"; proposition q: "equation 2x2-4x + M = 0 has no real root". If P ∧ q is false and P ∨ q is true, the value range of real number m is obtained
- 10. It is known that the square of ellipse x plus the square of 4Y is equal to 1 and the line y is equal to x plus M. when the line and ellipse have a common point, the value range of real number m is calculated? Don't you understand?
- 11. It is known that the eccentricity of ellipse C: x2 / A2 + Y2 / B2 = 1 is root sign 3 / 2, the straight line X-Y + 1 = 0 passes through the upper vertex of ellipse C, the straight line x = - 1 intersects the ellipse at two points a and B, P is different from the ellipse at any point a and B, the straight line AP BP intersects the straight line L: x = - 4 at two points Q, R, 1 respectively, to solve the equation of ellipse C, 2 to prove that the vector OQ × vector or is a fixed value
- 12. As shown in the figure, in the right angle trapezoid oabc, the coordinates of OA ‖ CB, a and B are a (15, 0), B (10, 12) respectively. The moving points P and Q start from O and B respectively. Point P moves along OA to terminal a at a speed of 2 units per second, and point Q moves along BC to C at a speed of 1 unit per second. When point P stops moving, point Q stops moving at the same time. The line segments OB and PQ intersect at point D, and make de ‖ when passing point D Let the motion time of PQ be t (in seconds). (1) when t is the value, the quadrilateral pabq is isosceles trapezoid, please write out the reasoning process; (2) when t = 2 seconds, calculate the area of trapezoid ofBC; (3) when t is the value, △ PQF is isosceles triangle? Please write down the reasoning process
- 13. In the plane rectangular coordinate system, O is the coordinate origin, oabc is a rectangle, point a coordinates (10,0) point C coordinates (0,4) point D is the midpoint of OA, point P is on the BC side When the triangle ODP is an isosceles triangle with waist length 5, the coordinate of point P is?
- 14. It is known that, as shown in the figure: in the plane rectangular coordinate system, O is the coordinate origin, the quadrilateral oabc is a rectangle, the coordinates of points a and C are a (10,0), C (0,4), point D is the midpoint of OA, and point P moves on the edge of BC. When △ ODP is an isosceles triangle with waist length of 5, the coordinate of point P is______ .
- 15. F (x) is an odd function defined on R, when x > = 0 is f (x) = x ^ 2, if for any x belonging to t to t + 2, the inequality f (x + T) > = 2F (x) is constant, the value range of T is obtained
- 16. If (X & # 178; - 2x + 1) of function f (x) = a is a decreasing function on (1,3), then the solution set of inequality log a (x + 1) > 0 about X is () A.{x|x>-1} B.{x|x>0} C.{x|x
- 17. The image of a certain function passes through the point (- 1,2), and the value of function y decreases with the increase of independent variable x. please write a function relation that meets the above conditions___ .
- 18. The value range of the independent variable in the function y = radical (x + 2) / X clearly indicates that the coefficient K of the inverse proportional function is not equal to 0. Why is the answer x > = - 2 and X ≠ 0 Value range of independent variable in function y = radical (x + 2) / X Why is the answer x > = - 2 and X ≠ 0 not x > - 2 and X ≠ 0
- 19. If (A & # 178; - 3a-2) + a of y = (A-4) x is a quadratic function, find the value of (1) a and the relation of (2) function
- 20. Since the independent variable x ≠ 0 and the function value y ≠ 0 in the inverse scale function y = K / X (K ≠ 0), its image is at the intersection of X axis and Y axis, that is, the two branches of hyperbola are infinitely close to 2, but never reach the coordinate axis. In addition, the two branches of hyperbola are symmetrical about coordinate 3, and about straight line y = x and straight line y = - x 4