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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______ .
(1) When od is the bottom of an isosceles triangle, P is the intersection of the vertical bisector of OD and CB, and op = PD ≠ 5; (2) when od is a waist of an isosceles triangle: ① if point O is the vertex of vertex, point P is the intersection of an arc with radius 5 and CB, and in the right angle △ OPC, CP = op2-oc2 = 52-42 = 3, then the coordinates of P are (3,4) In the right angle △ PDM, PM = pd2-dm2 = 3, when p is on the left side of M, CP = 5-3 = 2, then the coordinates of P are (2,4); when p is on the right side of M, CP = 5 + 3 = 8, then the coordinates of P are (8,4). So the coordinates of P are: (3,4) or (2,4) or (8,4) ）Or (8, 4)
The focus of hyperbola C2 ellipse C1 is the vertex, the vertex is the focus, and B is any point on the first quadrant of hyperbola
A f is the right vertex of the ellipse and E = (1 / 2) ∧ 0.5 of the left focus ellipse
B=c
Hyperbola C2, the focus of ellipse C1 is the vertex, the vertex is the focus, and B is any point on the first quadrant of hyperbola. Let's ask whether there is a constant λ such that the angle BAF = λ BFA is constant. If there is, find the value of λ
Let B (x1, Y1)
A (x, 0) f (- sqrt (2) / 2x, 0) (sqrt root) can be easily obtained
But when I make a vertical line from B to x, I'm depressed,
Tan (angle BAF) = Y1 / (x-x1)
Tan (angle BFA) = Y1 / (x1 + sqrt (2) / 2x)
How can we not find that the two angles are proportional to each other
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 (20,0), C (0,8), point D is the midpoint of OA, and point P moves on the side of BC. When △ ODP is an isosceles triangle with waist length of 10, the coordinate of point P is___ .
∵ a (20,0), C (0,8), the quadrilateral oabc is a rectangle, D is the midpoint of OA, ∵ OC = 8, OD = 10, ∵ OCB = ∵ cod = 90 °, ① OP = od = 10, obtained from Pythagorean theorem: CP = 102-82 = 6, that is, the coordinates of P are (6,8); ② DP = od = 10, PM ⊥ OA over P to m, then PM = OC = 8, obtained from Pythagorean theorem
Through the left focus of the ellipse x ^ 2 / 9 + y ^ 2 = 1, make a straight line and the ellipse intersect at two points ab. if the tension of the chord AB is equal to the length of the minor axis, find the linear equation
a=3,b=1,c=2√2.F1(-2√2,0),
Let the linear equation be y = K (x + 2 √ 2)
It is associated with elliptic equations: (1 + 9K & # 178;) x & # 178; + 36 (√ 2) K & # 178; X + 72K & # 178; - 9 = 0. ⊿ = 36 (K & # 178; + 1)
From the chord length formula: √ (1 + K & # 178;) * [(√ ⊿) / (1 + 9K & # 178;)] = 2
∴k=±√3/3.
The linear equation y = ± (√ 3 / 3 (x + 2 √ 2)
As shown in the figure, in the rectangular coordinate system, the quadrilateral oabc is a rectangular trapezoid. BC ‖ OA, ⊙ P is tangent to OA, OC and BC at points e, D and B respectively, and intersects AB at point F. given a (2,0), B (1,2), then Tan ∠ FDE=______ .
Connect Pb, PE. ∵ P with OA, BC respectively, tangent to points e, B, ∵ Pb ⊥ BC, PE ⊥ OA, ∵ BC ∥ OA, ∵ B, P, E in a straight line, ∵ a (2,0), B (1,2), ∵ AE = 1, be = 2, ∵ Tan ∠ Abe = aebe = 12, ∵ ∠ EDF = ∠ Abe, ∵ Tan ∠ FDE = 12
AB is the chord of the left focus f passing through the ellipse x ^ 2 / 5 + y ^ 2 / 4 = 1. If the line L intersects the ellipse at two points of AB, if the chord length of AB is (16 √ 5) / 9, the equation of line L is obtained
This is the first time that is 4x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\35;178; + 10K & #178; X + 5K & #178; - 20 = 0
As shown in the figure, in the rectangular coordinate system, the coordinates of the fixed point B of the rectangular oabc are (8,6), and the straight line y = 2 / 3 + m will exactly hold OA
The straight line y = 2 / 3 + m just divides the rectangular oabc into two parts with equal area, and calculates the value of M
Wrong number just now
A straight line passing through the center of a rectangle can divide the area of the rectangle equally
So the line must pass (4,3)
Substituting the line y = (2 / 3) x + m, M = 1 / 3 is obtained
Through the left focus of the ellipse x2 + 2Y2 = 4, make a straight line L with an inclination angle of 30 degrees, intersect the ellipse at two points a and B, find the equation of the straight line L, and the coordinates of the midpoint of the length ab of the chord ab
The focus of the (2,0) line is the slope of the line. The slope of the straight line is k = Tan 30 ° = √ 3 / 3 ᦉ; the linear equation is: y = {3 / 3 (x - √ 2) and ellipse x-178; (4 + 2Y ᦉ 178; + 2,0) the slope of the line is the slope of the line, and the slope of the straight line is k = Tan, 30 ° 30 ° 30 ° = √ 3 / 3 / 3 / 3 / 3 / 3 {3 / 3 / 3 (x - {2 / 3 (x-3) (x-178) (x-3 (x-3 / 3 (x-3) (x-3) (178)); the equation is: the equation is: y equation is: y equation is: y = {3 / 5x5x &, that is: y-5x &; 178; 178; 178; - 4; it's not easy
In the plane rectangular coordinate system, O is the origin of the coordinate, the quadrilateral oabc is the rectangle, the coordinate of point a is (10,0) C (0,4), and point D is the midpoint of OA, P is
When △ ODP is an isosceles triangle with waist length 5, the coordinates of point P are
When △ ODP is an isosceles triangle with waist length of 5
OP=OD=5
∵ OC=4
The OCP is RT △
∴ CP=3
The coordinates of point P are (3,4)
If we pass through the left focus of the ellipse x ^ 2 + 2Y ^ 2 = 4 and make a chord AB with an inclination angle of π / 3, then the length of the chord AB is
X ^ 2 + 2Y ^ 2 = 4x ^ 2 / 4 + y ^ 2 / 2 = 1A ^ 2 = 4, B ^ 2 = 2, C ^ 2 = a ^ 2-B ^ 2 = 2, left focus coordinate: (- √ 2,0) tilt angle is π / 3, KAB = Tan π / 3 = √ 3AB equation is: y = √ 3 (x + √ 2) substitute: x ^ 2 + 2Y ^ 2 = 4 get: x ^ 2 + 2 (√ 3 (x + √ 2)) ^ 2 = 47x ^ 2 + 12 √ 2x + 8 = 0x1 + x2 = - 12 √ 2 / 7, x1x2 = 8 / 7 (x1
RELATED INFORMATIONS
- 1. 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?
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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______ .
- 8. 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
- 9. When the value of the first-order function y = (4 + 2m) x + M-4 is known, the image passes through one three four quadrants
- 10. Find the coordinates of the intersection point of the image of the function y = 2x + 1 and y = - 3x-4
- 11. 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
- 12. 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
- 13. 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___ .
- 14. 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
- 15. 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
- 16. 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
- 17. Given y = (m ^ 2-m-2) x ^ m ^ 2-2m-1, when m is, y is the inverse proportion function of X
- 18. If the mean value + 1 of the absolute value power of K of polynomial 4xy - 1 / 5 (K-2) y is a cubic binomial, then K=
- 19. Given the function f (x) = loga (AX-1) (a > 0, a ≠ 1); (1) find the domain of definition of function f (x); (2) discuss the monotonicity of function f (x)
- 20. The image of X-2 power + 1 (a > 0, a is not equal to 1) of function y = a must pass through the point?