Let p be a moving point on hyperbola x24-y2 = 1, o be the origin of coordinates, and m be the midpoint of line OP, then the trajectory equation of point m is______ ．

Let p be a moving point on hyperbola x24-y2 = 1, o be the origin of coordinates, and m be the midpoint of line OP, then the trajectory equation of point m is______ ．

Let m (x, y), then p (2x, 2Y) be substituted into hyperbolic equation to obtain x2-4y2 = 1, which is the trajectory equation of point m x2-4y2 = 1. Answer: x2-4y2 = 1

Given the fixed point m (0, - 1), the moving point P moves on the curve y = 2x ^ 2 + 1, the trajectory equation of the midpoint n of the line MP is obtained,

Let n (x, y) be the midpoint
be
The coordinates of point P are (2x-0,2y + 1), i.e. (2x, 2Y + 1)
And point P is on the curve y = 2x & # 178; + 1
therefore
2y+1=2(2x)²+1
2y=8x²
The trajectory equation is as follows
y=4x²

A mathematical problem is known that a (0.5, root sign 3 / 2, P and Q are two moving points on the circle x2 + y2 = 5, AP is perpendicular to AQ, then what is the maximum value of PQ?
A 4
B 2 radical 3
C 3 radical 2
D2 radical 5
The coordinates of point a are: (1 / 2, root 2 / 3), not root (2 / 3)
The process should be as detailed as possible, thank you

Answer: A, AP is perpendicular to AQ, and the maximum value of PQ is equal to any point a 'on the circle with o as the center and OA as the radius. Take p' and Q 'on the known circle to make a' p 'perpendicular to a' Q ', then the maximum value of P' Q 'is equal to the maximum value of PQ. Take a' (root 2 / 2, root 2 / 2), and P 'Q' = 4 can be obtained

As shown in the figure, AB is the chord of ⊙ o, and the radius od intersects AB at point C. AC: CB = 1:2, DC = 2, CO = 3 are known
There's no picture. Let's draw by ourselves
Maybe the picture is wrong

If do is extended and the intersection circle O is at point E, then OE = OC = 5 and CE = 8
Let AC = k, then BC = 2K
According to the intersection chord theorem
2k²=2*8=16
k=2√2
∴AB=3k=6√2
Make of ⊥ AB at point F
Then AF = 3 √ 2
∵OA=5
According to Pythagorean theorem, of = √ 7
That is, the chord center distance of AB is √ 7

It is known that AB is the chord (not the diameter) of the circle O. from any point of the circle, make the chord CD perpendicular to AB, make the angular bisector of the angle OCD intersect the circle at P, and connect PA and Pb

Certification:
Link Op
OC = OP = = > angle OCP = angle OPC
PC bisector angle OCD = = > angle OCP = angle PCD
So, angle OPC = angle PCD = = > OP parallel CD
CD vertical AB,
So OP is vertical ab
So, PA = Pb

It is known that AB is the chord of ⊙ O. from any point of the circle, make the chord CD ⊥ AB, make the bisector of ∠ OCD ⊙ o at P, connect PA and Pb, and prove that PA = Pb

Proof: because OP is the bisector of angle OCD,
So angle DCP = angle OCP,
And because OC = OP,
So angle OCP = angle OPC,
So angle DCP = angle OPC,
So CD is parallel to Op,
And because CD is vertical to AB, OP is vertical to ab,
So arc AP equals arc BP,
So PA = Pb

Let's know that AB is the chord of ⊙ o, and choose any point on the circle. Because of the chord CD ⊥ AB, make the bisector of ⊙ OCD intersect ⊙ o at P, connect PA and Pb, and prove that PA = Pb

Let AB and CD intersect at h, connect Po, intersect AB at g, extend Co, intersect ⊙ o at e, connect PD, PE and de. because PC bisects ⊙ DCE, then ed = PD, then ⊙ EOP = ⊙ DOP, it is easy to prove op ⊥ ed. because ⊙ CDE = 90 °, so op ∥ CD, because CD ⊥ AB, so Po ⊥ AB, so ⊥ PGB = ⊙ PGA = 90 °. Because Po is ⊙ o radius, so

As shown in the figure, AB is a diameter of circle O, which bisects circle O into upper and lower semicircles. From a point C on the upper semicircle, CD is perpendicular to AB, and the bisector of angle OCD intersects circle O at p. when point C moves on the upper semicircle (excluding two points a and B), does the position of point P change? Why?

Link op
Because OP equals OC
So the angle OCP is equal to the angle OPC
Because CP is the angular bisector of angular OCD
So the angle OCP is equal to the angle DCP
So the angle OPC is equal to the angle DCP
So CD is parallel to Op
Because CD is perpendicular to ab
So OP is perpendicular to ab
So the position of point P remains the same

AB is the diameter of the circle O, C is a moving point on the circle (not coincident with a and b), crossing point C as CD ⊥ AB, intersecting circle O to D, intersecting AB to F, intersecting bisector o to P
(1) Does the position of point P change with the position of point C? Please explain why
(2) Change diameter AB to chord AB, other conditions remain unchanged, whether the conclusion has changed? Please explain the reason

(1) Unchanged
Connect OC CD PC because CP is the bisector of angle OCD, so angle OCP = angle DCP and because OC
OP is the radius of the circle, so the angle OCP = angle OPC, so the angle DCP = angle OPC, so CD / / PC
And because CD ⊥ AB AB is a fixed point, so Po ⊥ ab
So p is not changed
(2) Invariant connection op
CP is the bisector of angle OCD, so angle OCP = angle DCP and because of OC
OP is the radius of the circle, so the angle OCP = angle OPC, so the angle DCP = angle OPC, so CD / / PC
And because CD ⊥ AB AB is a fixed point, so Po ⊥ ab
So p is not changed

As shown in the figure, in ⊙ o, AB is the chord of ⊙ o, C and D are two points on the straight line AB, and AC = BD. it is proved that △ OCD is an isosceles triangle

It is proved that: (1) when crossing point O, make om ⊥ AB, the perpendicular foot is m; ∵ om ⊥ AB, ∵ am = BM, ∵ AC = BD, ∵ cm = DM, and ∵ om ⊥ AB, ∵ OC = OD, ∵ OCD are isosceles triangles. (2) connect OA, OB; ∵ OA = ob, ∩ OAB = ∠ oba, ≌ CBO ≌ Dao, ≌ OC = OD, ∩ OCD are isosceles triangles. (3) (the same as above) (2) connect OA, ob, ∩ Cao = ∠ dbo, In addition, AC = BD, Cao, dbo and OCD are isosceles triangles