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

(1) As shown in the figure, if BG ⊥ OA is made to G through B, then AB = BG2 + Ga2 = 122 + (15 − 10) 2 = 169 = 13. If QH ⊥ OA is made to h through Q, then QP = QH2 + pH2 = 122 + (10 − t − 2t) 2 = 144 + (10 − 3T) 2. To make the quadrilateral pabq an isosceles trapezoid, then AB = QP, that is 144 + (10 − 3T) 2 = 13
If a straight line L passes through the left focus F of the ellipse x2 / 9 + y2 = 1, if the length of the chord AB passing through F is equal to the length of the minor axis, calculate AB to obtain the inclination angle
Let a (x0, Y0) B (x0, Y0), so AB = AF + BF = a + ex0 + A + EX1 = 2A + e (x0 + x1) = 6 + 2, radical 2 / 3 (x0 + x1) = 1ab, the linear equation is y = K (x + C), and substitute it into the elliptic equation to eliminate y to get (9K ^ 2 + 1) x ^ 2 + 18K ^ 2cx + 9K ^ 2C ^ 2-9 = 0, so x0 + X1 = - 18K ^ 2C / (9K ^ 2 + 1), and C = 2, radical 2, so k = ± radical 15 / 3
The coordinate of vertex B of rhombic oabc on Y-axis is - 3,2. If the image of inverse scale function y = K / x [x > 0] passes through point a, the value range of X is calculated when 1 < Y > 3
Modification: when 1 < y < 3, calculate the value range of X
diamond
O(0,0)
B(0,4)
A(3,2)
K=6
Two
Let f be the right focus of the ellipse x2 / 25 + Y2 / 9 = 1 and ab be the chord passing through the origin?
If the coordinate of point a is (x, y), then the coordinate of point B is (- x, - y). From the ellipse range, we know that - 3 ≤ y ≤ 3. The major axis of ellipse divides △ ABF into △ AOF and △ BOF (o is the origin of coordinate). Then s △ ABF = s △ AOF + s △ BOF = of Y0 + of Y0 = of Y0 = 4, Y0 ≤ 12
As shown in the figure, oabc is a rectangular piece of paper placed in the plane rectangular coordinate system, O is the origin, point a is on the positive half axis of X axis, point C is on the positive half axis of Y axis, OA = 10, OC = 8, take a point D on OC side, fold the paper along ad, so that point O falls on point E on BC side, then the coordinate of point D is______ .
In RT △ Abe, ab = 8, AE = 10, be = AE2 − AB2 = 6, CE = bc-be = 4, let od = x, then de = x, DC = 8-x, in RT △ CDE, ∵ de2 =..., and let od = x, then de = x, DC = 8-x
What are the lengths of the longest and shortest strings passing through the focus of the ellipse x2 / 4 + Y2 / 3 = 1?
The longest chord of the focus of the ellipse x2 / 4 + Y2 / 3 = 1 is the major axis, the length is 2A = 4, and the shortest chord is the path. Through the focus, make a straight line perpendicular to the X axis, intersect the ellipse at a, B, then AB is called the path of the ellipse, ∵ C & # 178; = A & # 178; - B & # 178; = 1, ∵ C = 1. Substituting x = 1 into x2 / 4 + Y2 / 3 = 1, we get y & # 178; = 9 / 4, | y | = 3 / 2, the path length is 2
As shown in the figure, in the plane rectangular coordinate system, the quadrilateral oabc is a square with side length of 2, and the vertices a and C are respectively on the positive half axis of X and Y axes. Point q is on the diagonal ob, and Qo = OC. Connect CQ and extend the intersection ab of CQ to point P. then the coordinate of point P is___ .
∵ quadrilateral oabc is a square with side length of 2, ∵ OA = OC = 2, OB = 22, ∵ Qo = OC, ∵ BQ = ob-oq = 22-2, ∥ OC, ∥ bpq ∥ OCQ, ∥ BPOC = bqoq, i.e., bp2 = 22-22, the solution is BP = 22-2, ∥ AP = ab-bp = 2 - (22-2) = 4-22, the coordinates of point P are (2, 4-22). So the answer is: (2, 4-22)
In the ellipse x2 / 9 + y2 = 1, find the chord length of the chord with right focus and 45 ° inclination angle
First, get the focus. Second, set a straight line L that passes through the focus and has a slope of 1. Third, get the sum and product of the abscissa of the intersection of the ellipse and the straight line. Finally, use the chord length formula. The field formula is chord length = √ 1 + K ^ 2 * lx2-x1. Because it's too cumbersome to connect the ellipse and the straight line, I'm sorry
There are many ways to do it
Chord length formula, polar coordinates, focal radius
Answer 0.8
2 / 5 you make full use of 45 degrees, do not set a linear equation, just set a Y Y > 0, substitute it with the focus relation x = C Y to get y, and then find the chord should be 2 / 5, you can calculate it yourself!
It is known that the eccentricity of ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0) is root 2 / 2, A1 and A2 are the left and right vertices of ellipse, b1b2
It is known that the eccentricity of ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0) is radical 2 / 2, A1 and A2 are the left and right vertices of ellipse, b1b2 is the upper and lower vertices of ellipse, and the area of quadrilateral A1A2B1B2 is 16 radical 2
(1) This is the case of C / a = C / a = 2 / 2 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\178; = 16 B & # 17
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

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