![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
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
(1) AC angle = - (45 ° + Θ), Mn angle = Θ - 90 °. & nbsp; AC ‖ Mn & nbsp; & nbsp; - (45 ° + Θ) = Θ - 90 °. & nbsp; & nbsp; Θ = 22 ° 30 ′ (2) & nbsp; △ omn ≌ Δ OMT (SAS) & nbsp; Mn = MT = CN + am & nbsp; - P = Ba + BC =
Given the ellipse x ^ 2 / 2 + y ^ 2 = 1 and the point B (0, - 2) passing through the left focus F1 and the straight line intersection ellipse of point B at C and D, the right focus of the two-point ellipse is F2, and the area of triangle CdF2 can be calculated
According to the meaning of the title, the focus of the ellipse is: F1 (- 1,0), F2 (1,0)
Let the line passing through the left focus F1 and point B be y = KX + B
Then: - K + B = 0, 0 + B = - 2
The solution is: k = - 2, B = - 2
The line passing through the left focus F1 and point B is y = - 2x-2
∵ the ellipse passing through the intersection of the left focus F1 and point b lies at two points c and D
The distance between the right focus F2 (1,0) and CD d = │ 2 × 1 + 1 × 0 + 2 │ / √ 2 ^ 2 + 1 ^ 2 = 4 / √ 5
∵ the ellipse passing through the intersection of the left focus F1 and point b lies at two points c and D
The solution is: x = (- 8 ± √ 10) / 9, y = - 2 (1 ± √ 10) / 9
∴│CD│=10√2 /9
S=1/2 × (10√2 /9) ×(4/√5)=4√10 /9
F1 (- 1,0) and B point can be solved, and the equation is 2x + y = - 2, which can be obtained by substituting it into elliptic equation
9y ^ 2 + 4Y = 4 only requires the absolute value of y1-y2
Easy to get is 4 / 3 root 10
So the area is 4 / 3 times the root 10
It is known that the inverse scale function y = 3-2m / X is within the image limit of the image, and Y decreases with the increase of X. the value range of the letter M is obtained
The solution consists of the inverse scale function y = 3-2m / X. within the image limit, y decreases with the increase of X
Then 3-2m ＞ 0
That is 2m < 3
M < 3 / 2
The center of the ellipse C is at the origin, the focus is on the x-axis, the eccentricity e = 1 / 2, and the coordinates of a vertex are (0, √ 3)
(1) Find the equation of ellipse C (2) the left focus of ellipse C is f, the right vertex is a, the line L: y = KX + m intersects with ellipse C at M and N, and the vector am * vector an = 0. Ask whether there is a real number L, so that s △ FMN = LS △ amn holds. If there is, find out the value of L
Elliptic equation x ^ 2 / a '^ 2 + y ^ 2 / b ^ 2 = 1, then B = √ 3,' = 2, vector known am * an = 0, am is perpendicular to an, and then m, N, X axis points must be located on both sides of such a hypothetical point. The coordinates (x1, Y1) of X axis m located below are not X axis, coordinates (X2, Y2), the line L intersects with X axis
Given the image intersection of the first-order functions y = - 5 / 4x + (2m + 10) / 4 and y = - 2 / 3x + m / 3 in the fourth quadrant, we can find the integer M
Y = - 5 / 4x + (2m + 10) / 4 and y = - 2 / 3x + m / 3,
X=(2m+30)/7,y=(m-20)/7.
The image is intersected in the fourth quadrant
X>0,Y0,
m-20
There are two cases: the intercept is on the positive half axis, and M / 3 > (2m + 10) / 4 > = 0 m has no solution
So take the second case
(2m+10)/4
In the plane rectangular coordinate system, the center of the ellipse is at the origin of the coordinate, its focus is on the x-axis, and its vertex is on the line x + 2y-2 = 0?
∵ the center of the ellipse is at the coordinate origin, its focus is on the X axis, and its vertex is on the line x + 2y-2 = 0
The intersection point of the straight line x + 2y-2 = 0 and the X axis is (2,0), that is, the top of the major axis of the ellipse is a = 2
The intersection point of the line x + 2y-2 = 0 and the Y axis is (0,1), that is, the top of the minor axis of the ellipse is b = 1
The standard equation of ellipse is X & # 178; / 4 + Y & # 178; = 1
Because the focus is on the x-axis, let x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, the vertex of x-axis is (a, 0), and the vertex of y-axis is (0, b).
And because the vertex of the ellipse is on the straight line, it must be on the intersection of the straight line and the coordinate axis, and X, y = 0 respectively, we get (0,1), (2,0). So a = 2, B = 1.
So x ^ 2 / 4 + y ^ 2 = 1. Thank you for your adoption!
In the plane rectangular coordinate system xoy, the vertex of the parabola C is at the origin, passing through the point a (2), the focus f (0.5, 0) is obtained first, and then the OA equation y = x is obtained, because it is vertical, so the slope is not good
Because the focus is on the x-axis, let x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, the vertex of x-axis is (a, 0), and the vertex of y-axis is (0, b).
And because the vertex of the ellipse is on the straight line, it must be on the intersection of the straight line and the coordinate axis, and X, y = 0 respectively, we get (0,1), (2,0). So a = 2, B = 1.
What is the value range of m when the image of a linear function y = (m-2) X-1 passes through two three four quadrants
Analysis of test questions: there are four kinds of images of linear function
① When, the image of the function passes through the first, second and third quadrants;
② When, the image of the function passes through the first, third and fourth quadrants;
③ When, the image of the function passes through the first, second and fourth quadrants;
④ When, the image of the function passes through the second, third and fourth quadrants,
∵ the image of a linear function y = (m-2) X-1 passes through two, three and four quadrants,
The solution is m-2 < 0
Are you satisfied with my answer? Give me a good comment, or you can continue to ask me
A (4,0) B (0,5) are the two vertices of x ^ 2 / 16 + y ^ 2 / 25 = 1 of the ellipse. C is a point in the first quadrant of the ellipse. Find the maximum value of the triangle ABC
No need
AB = 41 under root
So only the maximum distance from C to AB is C (x, y)
By two-point formula lAB:5X+4Y-20=0
D = | 5x + 4y-20 | / 41
10> 0 Y > 0 5x + 4Y is the maximum
(5X+4Y)^2
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______ .
According to the meaning of the question, we get 2-m ＜ 0 and m ＞ 0, then we get m ＞ 2
Given that the ellipse is 25 / x square + 16 / y square = 1, the vertices B and C of the triangle ABC coincide with the two focuses of the ellipse, and the point a moves on the ellipse, try to find the trajectory equation of the center of gravity g of the triangle ABC
B (- 3,0) C (3,0) let a (x0, Y0)
Center of gravity g (x, y)
Triangle center of gravity formula x = (- 3 + 3 + x0) / 3 = x0 / 3 x0 = 3x
y=y0/3 y0=3y
Point a moves on the ellipse x0 ^ 2 / 25 + Y0 ^ 2 / 16 = 1
9x^2/25+9y^2/16=1
The center of gravity divides the center line into two parts of 2:1.
Ao is the middle line, G is the center, that is OA: go = 3:1
That is, if G (x, y) is set, then a coordinate is (3x, 3Y)
If you bring it into the ellipse, that's the trajectory of G.
RELATED INFORMATIONS
- 1. 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______ .
- 2. 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
- 3. When the value of the first-order function y = (4 + 2m) x + M-4 is known, the image passes through one three four quadrants
- 4. Find the coordinates of the intersection point of the image of the function y = 2x + 1 and y = - 3x-4
- 5. 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
- 6. 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
- 7. 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
- 8. 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?
- 9. It is known that the ellipse G: X6 ^ 2 / 4 + y ^ 2 = 1. Through the point (m, 0), make the tangent l of the circle x ^ 2 + y ^ 2 = 1 and intersect the ellipse g at two points a and B It is known that the ellipse G: X6 ^ 2 / 4 + y ^ 2 = 1. The tangent l of circle x ^ 2 + y ^ 2 = 1 intersects the ellipse g at two points a and B through the point (m, 0). (1) find out the focus coordinate and eccentricity of the ellipse g. (2) express labl as a function of M, and find out the maximum value of labl
- 10. From the point a (- 1,4), make the tangent l of the square of the circle (X-2) + (Y-3) = 1, and find the tangent length From point a (- 1,4) make the tangent l of the square of the circle (X-2) + (Y-3) = 1
- 11. 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
- 12. 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
- 13. 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
- 14. 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
- 15. 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?
- 16. 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______ .
- 17. 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
- 18. 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
- 19. 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___ .
- 20. 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