Given that the angle a = 60, P and Q are the moving points on both sides of angle a, if the sum of the lengths of AP and AQ is the fixed value 4, the minimum PQ of the line segment is obtained

Given that the angle a = 60, P and Q are the moving points on both sides of angle a, if the sum of the lengths of AP and AQ is the fixed value 4, the minimum PQ of the line segment is obtained

Let p be the perpendicular of AQ to m, and let AP = X,
According to AP, the sum of AQ length is fixed value 4, AQ = 4-x
According to the cosine theorem,
PQ^2=x^2+(4-x)^2-2x(4-x)*cos60=3x^2-12x+16
The minimum value is when AP = x = 2, AQ = 2, PQ minimum value is 2

Circle a: (X-2) ^ 2 + y ^ 2 = 1, parabola C: y = - 1 / 4x ^ 2, cross the focus F of parabola as a straight line at two points m and n,
Then the G locus equation of the midpoint P of the chord Mn is?

MN⊥AP
P is on a circle with diameter AF
A (2,0), f (0, - 1), the trajectory equation is x (X-2) + y (y + 1) = 0, that is, x ^ 2 + y ^ 2-2x + y = 0 (the part in circle a)

When the line L passes through the point P (- 4.3) and intersects the x-axis and y-axis at two points a and B respectively, and AP / Pb = 3 / 5, the equation of line L is obtained

Let the equation of l be y = KX + B, make X and Y axes perpendicular through P, and intersect X and Y axes at C and D respectively,
It is known that: 3 = k * (- 4) + B
From the known two: according to the similar triangle principle: BD / PC = (B-3) / 3 = 5 / 3, the solution is b = 8, k = 4 / 5

The line L passes through the point P (- 5 - 4) and intersects with the x-axis and y-axis respectively at two points a and B, where | AP |: | ab | = 3:5
It's AP BP|

|AP|:|BP|=3:5,
λ = vector AP: vector Pb = ± 3 / 5
Let a (a, 0), B (0, b),
According to the coordinate formula of fixed proportion point, the following formula can be obtained
-5=(a±3/5×0)/(1±3/5),
A = - 8 or - 2
-4=(0=±3/5·b)/(1±3/5),
B = - 32 / 3 or 8 / 3
From the intercept formula of linear equation,
The linear equation is
4X + 3Y + 32 = 0 or 4x-3y + 8 = 0

In circle 0, AB is a chord, C and D are two points on a straight line AB, and AC = BD

Thinking route: & nbsp;
1. OA, OB are the radius of O, and △ OAB is isosceles △ and OA = ob, ∠ a = ∠ B, & nbsp;
2. AC = BD (known) so △ OAC ≌ △ OBD (both sides, equal angle) & nbsp;
3. Since △ OAC ≌ △ OBD, OC = OD (congruent △ corresponding edges are equal) & nbsp;
4. Because OC = OD, △ OCD is isosceles;
This is the line of reasoning;
Liou 848, you have to prove the similarity, you have to give reasons

As shown in Figure 2, in the circle O, AB is a chord, C and D are two points on the straight line AB, and AC = BD. it is proved that △ OCD is an isosceles triangle

Connecting OA and ob, we can prove that the triangle OAC and OBD are congruent. Using the edges and corners, we can get OC = OD, so the triangle OCD is isosceles triangle

Point a is a sextant point on the circle, point B is the midpoint of arc an, and point P is a moving point on radius on. If the radius of the circle is 1, the minimum value of AP + BP is obtained
thank

Excuse me: where is n? Is n the sextant adjacent to a?

As shown in the figure, the known point a is a trisection point on the semicircle with Mn as the diameter, point B is the midpoint of an, and point P is the point on the radius on. If the radius of ⊙ o is l, then the minimum value of AP + BP is ()
A. 2B. 2C. 3D. 52

Make a symmetric point a 'of point a about Mn, connect a' B, intersect Mn at point P, then PA + Pb is the smallest, connect OA ', AA', ob, ∵ point a and a 'are symmetric about Mn, point a is a trisection point on semicircle, ∵ a' on = ∠ AON = 60 °, PA = PA ', ∵ point B is the midpoint of arc an ^, ∵ Bon = 30 °, ∵ a' O

Point a is a trisection point on the semicircle, point B is the midpoint of arc an, point P is a moving point on the diameter Mn, and the radius of circle O is 1. Find the minimum value of AP + BP

Find the symmetric point a 'of a with respect to Mn
Then AP = a'p, a'B is also a diameter of the circle (because angle a'om = angle AOM, angle a'om + angle BM = 180)
AP+BP=A'P+BP

Circle O diameter Mn Finding the minimum value of AP + BP
The diameter of the circle is Mn, the third point of the upper semicircle is a (in the right semicircle), B is the midpoint of an arc, and P is any moving point on on. Connect AP and BP, and find the minimum value of AP + BP
Radius is 1, the answer is not very clear

This is to connect a line so that it can form an isosceles triangle and AP + BP can be on the same line, so that it is minimum