As shown in the figure, the vertices a, D and C of oabc and ADEF are on the coordinate axis, the point F is on the line AB, and the points B and E are on the function As shown in the figure, the vertices a, D and C of oabc and ADEF are on the coordinate axis, the point F is on the line AB, and the points B and E are on the function y = 1 / X (x)

As shown in the figure, the vertices a, D and C of oabc and ADEF are on the coordinate axis, the point F is on the line AB, and the points B and E are on the function As shown in the figure, the vertices a, D and C of oabc and ADEF are on the coordinate axis, the point F is on the line AB, and the points B and E are on the function y = 1 / X (x)

Let oabc's side length be a and ADEF's side length be B
A(a,0) B(a,a) C(0,a) E(a+b,b) F(a,b)
B. E on y = 1 / X
Then a = 1 / A, (a > 0)
So, a = 1
At the same time, B = 1 / (1 + b) (b > 0)
b=-1/2+√ 5/2
The coordinates of point e are (1 / 2 + √ 5 / 2, - 1 / 2 + √ 5 / 2)

As shown in the figure, vertices a, D and C of oabc and ADEF are on the coordinate axis, points f are on AB, and points B and E are on the image of function y = 4x (X & gt; 0)
(1) Find the area of the square oabc; (2) find the coordinates of point E

(1) Let B coordinate be (x, y), ∵ xy = 4, and the oabc area be 4; (2) let ed = y, then od = 2 + y. from Y (2 + y) = 4, Y2 + 2y-4 = 0, y = - 1 ± 5 (y = - 1-5), e (1 + 5, - 1 + 5)

As shown in the figure, the vertex ADC of the square oabc ADEF is on the coordinate axis, the point F is on AB, and the point B e is on the image with the function y = 4 / X (x greater than 0), then the point e coordinates

Let B (x0, Y0), then a (x0,0), C (0, Y0)
x0=y0
x0*y0=1
So we have x0 = Y0 = 1, that is B (1,1);
Let e (x1, Y1), D (x1,0), f (1, Y1)
x1-1=y1
x1*y1=1
x=√5+1 y=√5+1
So it's (√ 5 + 1, √ 5-1)

As shown in the figure, the vertices a, D, C of the square oabc and ADEF are on the coordinate axis, the point F is on AB, and the points B and E are on the image of the function y = 1x (x > 0), then the abscissa of the point E is______ ．

Let the coordinates of point B be (a, a) and a = 1a, and the solution is a = 1, that is, B (1, 1). Let the coordinates of point e be (1 + B, b), and E be on the image of function y = 1 x (x > 0); (1 + b) · B = 1, and the solution is b = − 1 ± 52, and b > 0, B = − 1 + 52, and the abscissa of point E = 1 + − 1 + 52 = 1 + 52