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In the Cartesian coordinate plane xOy, given the point A (3,2), the point B moves on the circle x^2+y^2=1, the moving point P satisfies the vector AP=vector PB, then the trajectory of the point P is? ... In the Cartesian coordinate plane xOy, given the point A (3,2), the point B moves on the circle x^2+y^2=1, the moving point P satisfies the vector AP=vector PB, then the trajectory of the point P is? Circular elliptic parabolic straight line
Let P (x, y), B (x1, y1)
Then (x-3, y-2)=(x1-x, y1-y)
So x1=2x-3, y1=2y-2
Because point B moves on circle x^2+y^2=1
So (2x-3)^2+(2y-2)^2=1
So point P's trajectory is a circle.
Let P (x, y), B (x1, y1)
Then (x-3, y-2)=(x1-x, y1-y)
So x1=2x-3, y1=2y-2
Because point B moves on circle x^2+y^2=1
So (2x-3)^2+(2y-2)^2=1
So point P's trajectory is round.
Curve C: y^2=x+1 and fixed point A (3,1), B is any point on curve C. If AP vector =2 times PB vector, when point B moves on curve C, Given the curve C: y^2=x+1 and the fixed point A (3,1), B is any point on the curve C. If the AP vector =2 times the PB vector, when point B moves on the curve C, find the trajectory equation of point P. Come on. Curve C: y^2=x+1 and fixed point A (3,1), B is any point on curve C. If the AP vector =2 times the PB vector, when point B moves on curve C, Given the curve C: y^2=x+1 and the fixed point A (3,1), B is any point on the curve C. If the AP vector =2 times the PB vector, when point B moves on the curve C, find the trajectory equation of point P. Come on.
Let P (x, y) B (xB, yB) x=(3+2xB)/(1+2) y=(1+2yB)/(1+2) xB=(3x-3)/2yB=(3y-1)/2 bring xB, yB into the parabola, and (3y-1)/2)^2=(3x-3)/2+1 sort out 9y^2-6y-6x+1=0, so the P-point trajectory equation is 9y^2-6y-6x+1=0...
Given the fixed point A (4,0), B is a moving point on the circle x^2+y^2=4, the point P satisfies the AP vector=2PB vector, and the trajectory equation of the point P Helps the respondent accurate answers
Let P (x, y), B (x1, y1)
As is known, the ratio λ=2 of the P component vector AB,
According to the formula of dividing point,
X=(4+2x1)/(1+2)=4/3+(2/3) x1
Y =(0+2y1)/(1+2)=(2/3) y1
Yes x1=(3x-4)/2 1, y1=3y/2 2
B is a moving point on circle x^2+y^2=4, so x1^2+y1^2=4 3
Substitute 1 2 into 3 and simplify to (3x-4)^2+9y ^2=16.
P.s This is an ellipse, which can not be converted into a standard form due to inconvenience. This is very simple
Given that A, B are two points on the ellipse x^2/4+y^2/3=1, AB is perpendicular to the x-axis, P is on the line segment AB, and vector AP* vector PB=1, find the trajectory equation of point P Given that A, B are two points on the ellipse x^2/4+y^2/3=1, AB is perpendicular to the x-axis, P is on the segment AB, and vector AP* vector PB=1, find the trajectory equation of point P It is known that A, B are two points on the ellipse x^2/4+y^2/3=1, AB is perpendicular to the x-axis, P is on the segment AB, and vector AP* vector PB=1, find the trajectory equation of point P
AB is vertical to the x-axis, so A and B are symmetric about the x-axis, let A (x1, y1), then B (x1,-y1), let P (x, y)
Then x=x1, AP=(0, y-y1), BP=(0, y+y1),
Since AP•BP=1,(y-y1)(y+y1)=1, i.e. y12=y2-1
Substituting the coordinates of A into the elliptic equation, we get x12/4+y12/3=1, i.e. x2/4+(y2-1)/3=1
Therefore, the trajectory equation of point P is 3x2/16+y2/4=1
A hyperbola passing through the origin has a focal point F (4,0),2a=2, and the trajectory equation of the center of the hyperbola is obtained.
0
Given that the straight line of the fixed point P (0,1) intersects the hyperbola x^2-y^2/4=1 at two points A and B, Q: if the midpoint of the straight line AB is M, find the trajectory equation of the M point The answer is 4x^2-y^2+y=0(y1). I just want to ask, how does the range of y come out. And why can't you take the equal sign.
The straight line l passing through the fixed point P (0,1) is: y=kx+1 substituted into the hyperbola, the result is 4x2-(kx+1)2=4, the result is (4-k2) x2-2kx-5=0 (1) set A, B points...
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