In the plane rectangular coordinate system, the hyperbola y = 12 / X and the straight line y = (3 / 4) x intersect at points a, B and OA = 5 If we want to find the point of intersection of the hyperbola, y = 3, x = 3 in the hyperbolic system, then we need to find out the coordinates of the first point in the hyperbola

In the plane rectangular coordinate system, the hyperbola y = 12 / X and the straight line y = (3 / 4) x intersect at points a, B and OA = 5 If we want to find the point of intersection of the hyperbola, y = 3, x = 3 in the hyperbolic system, then we need to find out the coordinates of the first point in the hyperbola

Intersection of y = 12 / X and y = (3 / 4) x a (4,3), B (- 4, - 3)
Let Q (x, 12 / x), where x > 0
If there is a point Q, then QA ^ 2 + QB ^ 2 = AB ^ 2
（x+4)^2+(12/x+3)^2+(x-4)^2+(12/x-3)^2=100
x^2=16 x^2=9
x=4 ,x=3
Q（4,3）,Q（3,4）

As shown in the figure, in the plane rectangular coordinate system, the straight line y = 1 / 2 x + 1 / 2 intersects point a with X axis, K intersects with hyperbola y = x part k at B, BC is perpendicular to X axis, OC = 2ao, solve hyperbolic analytic formula

Because the first function y = x / 2 + 1 / 2 intersects the x-axis and a point, so a (- 1,0), because OC = 2oa, so C (2,0), because BC ⊥ X axis tangent B on the first function, so B (2,3 / 2) because B is on the inverse function, so it can be y = 3 / X

As shown in the figure, in the plane rectangular coordinate system xoy, a branch of the straight line OA and hyperbola intersects at point a (2,2). After finding out the relationship between the straight line and hyperbola, answer the following question (2) After the straight line OA is translated up 3 units, the straight line and hyperbola intersect at B and C, and the area of △ BOC is calculated?

(1) Let the analytic formula of straight line be y = MX; let the analytic formula of hyperbola be y = KX
Then 2m = 2, M = 1; k = 2 × 2 = 4
The analytic formula of function of straight line OA is y = X;
The analytic formula of hyperbola is y = 4x
(2) After the straight line OA is translated up 3 units, the analytical formula of line CD is y = x + 3
According to the meaning of the title, we get
{y=x+3y=4x,
The solution is {x = 1y = 4 or {x = - 4Y = - 1
The intersection point C (1,4), D (- 4, - 1)
(3) Let the intersection of line CD and Y axis be e, then point E (0,3)
∴S△COD=S△COE+S△EOD= 3×12+3×42=7.5．
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I immediately regard [the intersection of straight line and hyperbola at B and C] as [the intersection of straight line and hyperbola at point C and D]. Doesn't it matter~

In the plane rectangular coordinate system xoy, two lines a and B parallel to the X and Y axes intersect at point a (3,4). Connecting OA, if there is a point P on the line a, then △ AOP is an isosceles triangle Three answers have been worked out -- (8,4) or (– 3,4) or (– 2,4), 50 classification. If there is no answer within half an hour, the question will be closed.

Midpoint D of OA (1.5,2)
The straight line passing through D and perpendicular to OA: Y-2 = - (3 / 4) (x = 1.5)
The intersection of this line and the straight line: y = 4 E (- 3.5 / 3,4)
For another point

As shown in the figure, in the plane rectangular coordinate system, the hyperbola y = K / X passes through a point a, ab ⊥ X axis in the first quadrant, the perpendicular foot is B, and the s triangle AOB = 2 If the line y = x + K passes through point a and intersects with X axis at point C, find s △ ABC

It's done. Look at the picture

As shown in the figure, in the plane rectangular coordinate system, the straight line y = 1 2x+1 2 intersects with X axis at point a and hyperbola y = K X intersects point B in the first quadrant, BC, X axis at point C, OC = 2ao

OC = 2ao,
∵ when y = 0, 1
2x+1
2 = 0, x = - 1,
The coordinates of point a are (- 1, 0),
∴OA=1．
And ∵ OC = 2oa,
∴OC=2，
The abscissa of point B is 2,
Substitute the line y = 1
2x+1
2, y = 3
2，
∴B（2，3
2）．
∵ point B is on the hyperbola,
∴k=xy=2×3
2=3，
The analytic formula of hyperbola is y = 3
x．

As shown in the figure, in the plane rectangular coordinate system, the coordinates of point a are (8,0), the coordinates of point B are (0,6), and C is the midpoint of line ab. can I ask if there is a point P on the X axis so that the triangle with P, a and C as the vertices is similar to △ AOB? If it exists, calculate the coordinates of point p; if not, explain the reason

The reasons are as follows: ∵ AOB = 90 °, OA = 8, OB = 6; ∵ AB = 10. ∵ C is the midpoint of line AB, ∵ AC = 5. ① if P corresponds to B, then ᦻ PAC ᦻ Bao, ∵ PA: Ba = AC: Ao,  AP = 254, ? OP = oa-ap = 74,  P (74, 0). ② or if P corresponds to o, then

As shown in the figure, in the plane rectangular coordinate system, the straight line y = - x + 1 intersects X axis at point a, and intersects Y axis at point B. (1) find the length of segment AB; (2) if point E is on AB, OE is vertical (3) under condition (2), make om perpendicular to EF and cross AB to m, try to determine the quantitative relationship among be, EM and am, and prove your conclusion

(1) Let x = 0, y = 1, then the coordinates of point B are (0,1), so ob = 1; if y = 0, - x + 1 = 0, then x = 1, the coordinates of point a are (1,0), so OA = 1,
The △ OAB is an isosceles right triangle,
ν AB = radical 2;
（2）∵OE⊥OF,
∴∠BOE=∠AOF,
And ∵ ob = OA, OE = of,
∴△BOE≌△AOF,
∴BE=AF,
ν AF + AE = be + AE = AB = radical 2;
(3) The quantitative relationship of be, EM and am is: am 2 + be 2 = me 2
Connect MF, as shown in the figure, ∵ OE ⊥ of, and OE = of,
The △ OEF is an isosceles right triangle,
∵OM⊥EF,
/ / OM is the vertical bisector of EF,
∴MF=ME,
And ∵ BOE ≌ △ AOF,
∴∠OAF=∠OBE=45°,
∴∠FAM=90°,
∴AM²+AF²=MF²,
∴AM²+BE²=ME²．

As shown in the figure, in the rectangular coordinate plane, the line AB is perpendicular to the y-axis, and the perpendicular foot is B, and ab = 2. If the line AB is turned over along the y-axis, and point a falls at point C, then the abscissa of point C is______ .

The abscissa of point C is - 2

In the plane rectangular coordinate system, the straight line AB and X axis intersect at point a, and Y axis intersect at point B, and the straight line OC: y = x intersects with point C, so as to calculate the area of angle OAC

(1) (1) y = - 2x + 12, y = X
\X = 4, y = 4, so C (4,4)
\Let y = 0, - 2x + 12 = 0, then x = 6,  a (6,0)
∴OA=6
∴S△OAC=1/2×6×4=12