Mathematical problem: introduce two tangent lines PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1 1, introduce two tangent PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1, and the tangent points are a, B, and ∠ APB = 60 degrees, then the trajectory equation of the moving point P is________ 2. Let P (x, y) be a moving point on the circle x ^ 2 + (Y-1) ^ 2 = 1. If the inequality x + y + C > 0 holds, then the value range of C_____________ 3. We know that the equation of the line where the edge ab of the equilateral △ ABC is located is x √ 3 + y = 0, and the coordinates of point C are (1, √ 3) 4. It is known that the circle M: x ^ 2 + (Y-2) ^ 2 = 1, q is the moving point on the x-axis, QA and QB tangent the circle m to two points a and B respectively (1) (liberal arts) if | ab | = 4 √ 2 / 3, find the equation of line MQ (Science) prove that the straight line AB always crosses a fixed point (2) Find the locus equation of midpoint P of moving string ab It's better to analyze it

Mathematical problem: introduce two tangent lines PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1 1, introduce two tangent PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1, and the tangent points are a, B, and ∠ APB = 60 degrees, then the trajectory equation of the moving point P is________ 2. Let P (x, y) be a moving point on the circle x ^ 2 + (Y-1) ^ 2 = 1. If the inequality x + y + C > 0 holds, then the value range of C_____________ 3. We know that the equation of the line where the edge ab of the equilateral △ ABC is located is x √ 3 + y = 0, and the coordinates of point C are (1, √ 3) 4. It is known that the circle M: x ^ 2 + (Y-2) ^ 2 = 1, q is the moving point on the x-axis, QA and QB tangent the circle m to two points a and B respectively (1) (liberal arts) if | ab | = 4 √ 2 / 3, find the equation of line MQ (Science) prove that the straight line AB always crosses a fixed point (2) Find the locus equation of midpoint P of moving string ab It's better to analyze it

1. Let P coordinate (x, y), center coordinate (0, 0, 0), radius 1, Op divide 0, M > √ 2-2-2, 3, the distance from C point to AB, H = | | | 3 + √ 3 | 3 | 3 | (1 + 3) = √ 3, | AC | = | BC | = 2 √ 3 / √ 3 = 3 = 2, set a coordinate (x1, Y1, Y1, Y1), two-point formula, (x1-1) ^ 2 + (- √ 3x1 - √ 3) ^ 2 = 4, X1 = 0, X2 = - 1, Y1 = 1, Y1 = 1, Y1 = 1, Y1 = 1, Y1 = 1, Y1 = 1, Y10, y2 = √ 3

Two tangent lines PA and Pb are introduced from the moving point P to the circle x2 + y2 = 1, and the tangent points are a and B respectively, ∠ APB = 60 °, then the trajectory equation of the moving point P is______ .

If the coordinates of point P are (x, y), then | Po|=
x2+y2
∵∠APB=60°
∴∠AP0=30°
∴|PO|=2|OB|=2
Qi
x2+y2=2
That is, X2 + y2 = 4
So the answer is: x2 + y2 = 4

From the moving point P to the circle x * x + y * y = 1, introduce two tangent lines PA, Pb; the tangent points are a, B; the angle APB = 60 degrees, then what is the trajectory equation of the moving point p; how to analyze it?

It's a circle with a concentric radius of 2
x*x+y*y=4

Two tangent lines PA and Pb are introduced from the moving point P to the circle x2 + y2 = 1, and the tangent points are a and B respectively, ∠ APB = 60 °, then the trajectory equation of the moving point P is______ .

If the coordinates of point P are (x, y), then | Po|=
x2+y2
∵∠APB=60°
∴∠AP0=30°
∴|PO|=2|OB|=2
Qi
x2+y2=2
That is, X2 + y2 = 4
So the answer is: x2 + y2 = 4

Given the vector OP = (2,1), OA = (1,7), OB = (5,1), let X be a point on the straight line AP (o is the coordinate origin), then what is the minimum value of XA * XB?

The equation of straight line AP is: y = - 6x + 13
Set point a coordinate (x, - 6x + 13)
xa=(1-x,6x-6)
xb=(5-x,6x-12)
xa*xb=(1-x)(5-x)+(6x-6)(6x-12)=37x^2+40x-77
The minimum value is - 3249 / 37 at x = - 20 / 37

Given that the vector OA = (1,1) ob = (2,3) a point P on the y-axis has a minimum AP * BP, then the coordinates of point P are

Solution of P (0, b)
Then AP = op-oa = (0, b) - (1,1) = (- 1, B-1)
BP=OP-OB=(0,b）-（2,3）=（-2,b-3）
AP * BP
=-1×（-2）+（b-1）（b-3）
=b^2-4b+3+2
=b^2-4b+5
=（b-2）^2+1
≥1
If and only if B = 2, the equal sign holds,
When B = 2, AP * BP has a minimum value of 1,
At this point, P (0,2)

How to prove that the op vector = OA vector + xob vector / 1 + X when the point a, B, P on a straight line satisfies the collinearity of AP and Pb and O is any point in space One of the conditions here is how to convert AP vector to XPB vector

OP vector = OA + AP = OA + Tab = OA + T (ob-oa) = (1-T) OA + Tob
Let t = x / (1 + x), then 1-T = 1 / (1 + x), so OP vector = OA vector + xob vector / 1 + X

Vector OA (3,0), vector ob (0,4) vector AP = m vector Pb m > 0 OP = x (vector OA / OA) + (vector ob / ob) find the minimum value of 1 / x + 3 / y)

OP = x (vector OA / OA) + y (vector ob / ob) = (x / 3) OA + (Y / 4) ob
because
AP = m vector Pb, M > 0
So p, a, B are collinear, so x / 3 + Y / 4 = 1, and x > 0, Y > 0
that
1/X+3/y=(1/X+3/y)(x/3+y/4)=1/3+(1/4)(y/x)+(x/y)+(3/4)
>=2√[(1/4)(y/x)(x/y)]+13/12=25/12
So the minimum is 25/12

Let a, P and B be collinear and the vector AP = t vector AB, t ∈ R, and O ∈ ab. prove that the vector OP = (1-T) vector OA + T vector ob Let a, P and B be collinear and the vector AP = t vector AB, t ∈ R, and O ∈ ab Verification: vector OP = (1-T) vector OA + T vector ob

Vector AP = t vector AB,
(OP-OA)=t(OB-OA)
OP=OA+tOB-tOA=(1-t)OA+tOB

Given the vector OP = (1-3) vector OA + 1 / 3 vector ob, then vector AP =? Vector ab Thanks for your help. Thank you

Vector OP = (1-1 / 3) vector OA + 1 / 3 * vector ob
=Vector OA + 1 / 3 (vector ob vector OA)
=Vector OA + 1 / 3 vector ab
transposition:
ν vector OP vector OA = 1 / 3 vector ab
ν vector AP = 1 / 3 vector ab