As shown in Figure 1, in the plane rectangular coordinate system, the straight line AB intersects X axis at point a, intersects Y axis at point B, and point C is the moving point on line ab (1) If ∠ OAB is 20 ° larger than oba, OC ⊥ AB, calculate the degree of ∠ AOC; (2) As shown in Fig. 2, am bisection ∠ Bao, BM bisection ∠ OBN. When point a moves on the negative half axis of X axis, does the value of ∠ AMB change? If it does not change, calculate the degree of ∠ AMB; if it changes, please explain the reason; (3) As shown in Fig. 3, if ∠ OAB = 45 ° and ∠ OPA = ∠ BPD, ∠ BDP = ∠ ODF, then the following two conclusions are obtained: Please point out that there is only one correct conclusion

As shown in Figure 1, in the plane rectangular coordinate system, the straight line AB intersects X axis at point a, intersects Y axis at point B, and point C is the moving point on line ab (1) If ∠ OAB is 20 ° larger than oba, OC ⊥ AB, calculate the degree of ∠ AOC; (2) As shown in Fig. 2, am bisection ∠ Bao, BM bisection ∠ OBN. When point a moves on the negative half axis of X axis, does the value of ∠ AMB change? If it does not change, calculate the degree of ∠ AMB; if it changes, please explain the reason; (3) As shown in Fig. 3, if ∠ OAB = 45 ° and ∠ OPA = ∠ BPD, ∠ BDP = ∠ ODF, then the following two conclusions are obtained: Please point out that there is only one correct conclusion

(1) ∵ AOB = 90 °, OAB is 20 ° larger than oba,
Qi
∠OAB−∠OBA＝20°
∠OAB+∠OBA＝90° ，
The solution is: ∠ oba = 35 °,
∵OC⊥AB，
∴∠OCA=∠AOB=90°，
∴∠AOC=∠OBA=35°；
(2) The AMB value does not change;
∵∠BAM=1
2∠BAO，∠ABM=∠ABO+∠OBM=∠ABO+1
2（∠AOB+∠BAO）=∠ABO+1
2（90°+∠BAO），
∴∠AMB=180°-（∠BAM+∠ABM）=180°-[1
2∠BAO+∠ABO+1
2（90°+∠BAO）]=45°；
(3) (2) DF ⊥ OP is correct;
∵∠OAB=45°，∠AOB=90°，
∴∠OAB=∠OBA=45°，
∵∠OPA=∠BPD，
∴∠PDB=∠PDB，
∵∠BDP=∠ODF，
∴∠AOP=∠ODF，
∵∠AOP+∠POD=90°，
∴∠ODF+∠POD=90°，
∴∠OED=90°，
∴DF⊥OP．

As shown in the figure, in the plane rectangular coordinate system xoy, the straight line AB and X axis intersect at point a, and Y axis intersect at point B, and OA = 3, ab = 5 As shown in the figure, in the plane rectangular coordinate system xoy, the straight line AB and X axis intersect at point a and Y axis intersect at point B, and OA = 3, ab = 5. Point P starts from point O and moves along OA at the speed of 1 unit per second to point a, and immediately returns along Ao at the original speed after reaching point a; point Q starts from point a and moves along AB at a speed of 1 unit per second to point B at a uniform speed, When point Q reaches point B, the point P will stop. The time for point P and Q to move is T seconds (T > 0) (1) Find the analytic formula of line ab; (2) In the process of point P moving from O to a, the functional relationship between the area s and t of △ Apq is obtained (it is not necessary to write out the range of T); (3) In the process of point e moving from B to o, the following problems are completed: ① Can the quadrilateral qbed become a right angle trapezoid? If so, ask for the value of T; if not, please explain the reason; ② When de passes through point O, please write the value of T (there must be a procedure)

(1) In RT △ AOB, OA = 3, ab = 5, OB = ab2-oa2 = 4. A (3,0), B (0,4) is obtained by Pythagorean theorem. Let the analytic formula of straight line AB be y = KX + B. {3K + B = 0b = 4. The analytic formula of line AB is y = - 43b = 4. The analytic formula of line AB is y = - 43x + 4; (2) as shown in Fig. 1, QF ⊥ Ao is made at point f. ∵ AQ = OP = t

As shown in the figure, in the plane rectangular coordinate system, the straight line AB intersects with X axis and Y axis at two points a (3,0) and B (0,3), As shown in the figure, in the plane rectangular coordinate system, the straight line AB intersects with X axis and Y axis respectively at a (3,0), B (0,3) two points, point C is a moving point on the line AB, passing through point C as CD ⊥ X axis at point D (1) Find the analytic formula of line ab; (2) If s trapezoid obcd = 4 root sign 3 / 3, calculate the coordinates of point C; (3) Whether there is a point P in the first quadrant, so that the triangle with P, O, B as the vertex is similar to △ oba. If there is, ask for the coordinates of all points P that meet the conditions; if not, please explain the reason

(1)
Let AB be y = KX + B, ∵ y = KX + B. through a, B two points ｛ 3K + B = 0; b = 3}, the solution {k = - 1, B = 3}
The analytic formula of AB is y = - x + 3
(2)
∵ OA = ob = 3, ∵ AOB = 90 °, CD ⊥ X axis  OAB = ∠ oba = 45 °, CD = ad, OD = 3-cd
Let CD be x (0 ≤ x < 3). (3 + x) (3-x) / 2 = 4 √ 3 / 3, the solution x = √ (81-24 √ 3) / 3
∴C｛3- √（81-24√3）/3,√（81-24√3）/3｝
(3)
existence
When OP is bevel edge, make PE ⊥ X axis
∵∠OBP=90° BP=OB=3 ,∴ OP=3√2 ,OE=3 ∴P1（3,3）
When BP is an oblique edge,
∵ ob = op ᙽ OP coincides with OA, P is on the x-axis

As shown in the figure, in the plane rectangular coordinate system, the straight line AB intersects with the x-axis and the y-axis at two points a (3,0), B (0, root 3), and point C is a line As shown in the figure, in the plane rectangular coordinate system, the straight line AB intersects with the x-axis and y-axis at two points a (3,0) and B (0, root 3), respectively. Point C is a moving point on line AB, and passing through point C makes CD ⊥ X-axis at point D (1) Whether there is a point P in the first quadrant, so that the triangle with P, O, B as the vertex is similar to △ oba. If there is, ask for the coordinates of all points P that meet the conditions; if not, please explain the reason

The △ oba is a right triangle, so △ POB must also be a right triangle
If I have a right angle to BOP, then p is on the X axis, which is in contradiction with P in the first quadrant in the title, so there are two cases,
That is, ∠ PbO = 90 ° and ∠ BPO = 90 °
① If ∠ PbO = 90 °, let P (x, √ 3), so √ 3 / x = 3 / √ 3 or √ 3 / x = √ 3 / 3, so x = 1 or 3,
So p (1, √ 3) or (√ 3, √ 3)
② If ∠ PbO = 90 ° then because of the similarity, the corresponding angles must be equal, so op ⊥ AB is on P, that is, the point P is on the straight line ab
BP / √ 3 = √ 3 / AB, because AB = 2 √ 3, so BP = √ 3 / 2, P (3 / 4,3 √ 3 / 4) can be obtained

In the plane rectangular coordinate system, the straight line AB intersects with x-axis and y-axis respectively at two points a (3,0) and B (0, root 3), and point C is a moving point on line ab Pass through point C as CD, perpendicular to X axis at D 1. Find the analytic formula of line ab 2. If s trapezoid obcd = (4 roots 3) / 3, find the coordinates of point C 3. Whether there is a point P in the first quadrant, so that the triangle with P, O, B as the vertex is similar to the triangle oba. If there is, ask for the coordinates of all points P that meet the conditions; if not, please explain the reason!

1. The analytic formula is y = - root of three * (x-3)
2. From (4 root sign 3) / 3 = [root 3 * (x-3) / 3 + root 3] * x, the abscissa of point C is equal to 2 roots 2, and the ordinate is obtained by the analytic formula
3. The point P coordinates are (3, root 3)

In the plane rectangular coordinate system xoy, we know the center of circle C (3.0) and pass through the point (- 1. - 3), and find the standard equation of circle C

Let the standard equation (x-3) 2 + y 2 = R
Again (- 1, - 3)
∴16+9=r²
r²=25
∴（X-3）²+Y²=25

In the plane rectangular coordinate system xoy, if the circle passes through point a (13, 0) with origin o as the center, and the straight line y = kx-3k + 4 and ⊙ o intersect B and C, then the minimum length of chord BC is___ .

∵ the line y = kx-3k + 4 = K (x-3) + 4, ᙽ K (x-3) = y-4, ? K has numerous values, ? x-3 = 0, y-4 = 0, x = 3, y = 4, ? the line must pass through point d (3,4), ? the shortest chord CB is the chord passing through point D and perpendicular to the diameter of the circle, ? the coordinate of point D is (3,4), ? od = 5, M with origin o as the center of the circle

In the plane rectangular coordinate system xoy, if the circle crossing point a (- 10,0) with the origin o as the center, and the straight line y = KX + 3k-4 and ⊙ o intersect at B and C, then the minimum length of chord BC is_____ . 10 root number 3 Before 20:30 today

According to the line y = kx-3k + 4 must pass through point d (3,4), the shortest chord CB is the chord passing through point D and perpendicular to the diameter of the circle, then the length of OD is calculated, and then the length of ob is calculated according to the circle crossing point a (- 10,0) with origin o as the center, and then the answer can be obtained by using Pythagorean theorem
∵ the line y = kx-3k + 4 must pass through point d (3,4),
The shortest chord CB is the chord passing through point D and perpendicular to the diameter of the circle,
∵ the coordinates of point D are (3,4),
∴OD=5,
∵ the circle crossing point a (- 10,0) with the origin o as the center,
The radius of the circle is 10,
∴OB=-10,
ν BD = 5 root number 3;
The minimum length of BC is 10 and 3

In the plane rectangular coordinate system xoy, if the circle passes through point a (13, 0) with origin o as the center, and the straight line y = kx-3k + 4 and ⊙ o intersect B and C, then the minimum length of chord BC is___ .

∵ the line y = kx-3k + 4 = K (x-3) + 4,
∴k（x-3）=y-4，
∵ K has countless values,
ν x-3 = 0, y-4 = 0, x = 3, y = 4,
The straight line must pass through point d (3, 4),
The shortest chord CB is the chord passing through point D and perpendicular to the diameter of the circle,
∵ the coordinates of point D are (3, 4),
∴OD=5，
∵ the circle passing point a (13,0) with the origin o as the center,
The radius of the circle is 13,
∴OB=13，
∴BD=12，
The minimum length of BC is 24;
So the answer is: 24

In the plane rectangular coordinate system, there is an ellipse with F1 (0, - radical 3) and F2 (0, radical 3) as the focus, and the eccentricity is half root 3 Let the part of ellipse in the first quadrant be curve C, the moving point P is on C, the intersection points of tangent line of C at point P and X, Y axis are respectively a, and vector om = OA vector + ob vector. The trace equation of M and the minimum value of OM vector module are obtained.

The ellipse with the focus of F1 (0, - radical 3) and F2 (0, radical 3) with eccentricity of half root 3
Obviously, a = 2, C = √ 3, B = 1,
The elliptic equation is x 2 / 4 + y 2 / 1 = 1;
The part of the ellipse in the first quadrant
Let point p be (x0, Y0)
Y '= - x0 (2 √ (4-x? 0)) is the slope of tangent line passing through point P
Y - Y0 = - x0 (2 √ (4-x? 0)) * (x - x0) is the tangent equation
Therefore, point a is (4 / x0,0), similarly, point B is (0,1 / Y0),
Om = OA vector + ob vector -- > m (4 / x0,1 / Y0);
Let x = 4 / x0, 1 / x = x0 / 4, and 1 / y = Y0
Because ellipses satisfy
x²0/4 + y²0/1 =1; (x0/4*2)² +y²0 =1;
-->(2 / x) 2 + (1 / y) 2 = 1 is the trajectory equation of M
M (4 / x0,1 / Y0); x 2 0 / 4 + y 2 0 / 1 = 1; x 0 = 2 cosa; y 0 = Sina; x 2 0 / 4 + y 2 0 / 1 = 1;
Om 2 = (2 / COSA) 2 + (1 / Sina) 2 = 2-T / T (1-T) (t = cos? A) = u (0, the maximum value u > = 1 / 2 can be obtained by the discriminant
|OM|>=√2/2