As shown in the figure, in the plane rectangular coordinate system, ob is on the x-axis, ∠ ABO = 90 ° and the coordinates of point a are (1,2). Rotate △ AOB 90 ° counterclockwise around point a, and the corresponding point C of point O just falls on a branch of hyperbola y = KX. (1) find the analytic formula of hyperbola. (2) the intersection of the straight line passing through point c y = - x + B and the hyperbola is e, and find the coordinates of point E and the area of △ EOC | Math enthusiast | Mathematical art

As shown in the figure, in the plane rectangular coordinate system, ob is on the x-axis, ∠ ABO = 90 ° and the coordinates of point a are (1,2). Rotate △ AOB 90 ° counterclockwise around point a, and the corresponding point C of point O just falls on a branch of hyperbola y = KX. (1) find the analytic formula of hyperbola. (2) the intersection of the straight line passing through point c y = - x + B and the hyperbola is e, and find the coordinates of point E and the area of △ EOC

(1) It can be seen from the rotation that C (3,1), substituting C (3,1) into y = KX, we can get the analytic formula of hyperbola of k = 3, ｜ is y = 3x; (2) substituting C (3,1) into y = - x + B, we can get the analytic formula of B = 4, ｜ straight line is y = - x + 4. ｜ - x + 4 = 3x, we can get the solution of X1 = 1, X2 = 3, ｜ e (1,3), ｜ s △ EOC = 3 × 3 − 12 × 1 × 3 − 12 × 2 × 2 = 4

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