If the moving point P (x0, Y0) moves on the curve y = 2x & # 178; + 1, find the trajectory equation of the midpoint of the line between P and point (0, - 1)

If the moving point P (x0, Y0) moves on the curve y = 2x & # 178; + 1, find the trajectory equation of the midpoint of the line between P and point (0, - 1)

Let the coordinates of the midpoint of the line be (a, b), then a = x0 / 2, B = (Y0 + 1) / 2, so x0 = 2A, Y0 = 2b-1
Because P is on the curve, so Y0 = 2x0 ^ 2 + 1, which is 2b-1 = 2 (2a) ^ 2 + 1. Simplify: B = 4A ^ 2 + 1, replace B with y, a with X, and the midpoint trajectory is y = 4x & # 178; + 1

If the moving point P (x1, Y1) moves on the curve y = 2x ^ 2 + 1, then the trajectory equation of the midpoint of the line between P and point (0, - 1) is obtained

Let Q (a, b) be the midpoint of the line between P and (0, - 1)
So the coordinates of point P are (2a. 2B + 1)
And because the point P is on the curve, it's brought in
8a^2+1=2b+1
So the trajectory equation of point q is y = 4x ^ 2

The moving point P moves on the circle C: x2 + y2 = 9, and the fixed point Q (5,0) is used to find the trajectory equation of the midpoint m of the line segment

If x2 + y2 = 9, y = √ 9-x2, let p be (x, y)
Then M is ((5 + x) / 2, Y / 2)
y/2=1/2√9-x2=1/2√(3-x)(x+3)=1/2√[8-(5+x)][ (5+x)-2]
Let m be (m, n), then n ^ 2 = (8-m) (m-2)
It is reduced to (m-5) ^ 2 + n ^ 2 = 9

Given the fixed point P (1,0), the moving point q is in the circle C (x + 1) ^ 2 + y ^ 2 = 1, the vertical bisector of PQ intersects the straight line CQ and the point m, then the trajectory of the moving point m is?

The center of circle C is C (- 1,0)
The intersection of PQ's vertical bisector line CQ and point m
∴ PM=MQ
Mq-mc = CQ
That is mq-mc = ± CQ
That is mp-mc = ± 1
The trajectory of M is a hyperbola with P and C as the focus and 1 as the real axis length
The equation is X & # 178; / (1 / 4) - Y & # 178; / (3 / 4) = 1

1. If we know the fixed point P (2,0), the moving point q is on the circle x ^ 2 + y ^ 2 = 9, and the vertical bisector of PQ intersects OQ at the point m, then the trajectory of the moving point m is?
2. Through the left vertex a of hyperbola M: x ^ 2 - (y ^ 2) / (b ^ 2) = 1, make a straight line L with slope 1. If l and two asymptotes of hyperbola m intersect at B and C respectively, and ∣ ab ∣ = ∣ BC ∣, then the eccentricity of hyperbola m is?

(1) Let m (x, y) Q (x0, Y0); the midpoint of PQ be n; 2. Write KPQ first; then write knm; 3. Because two straight lines are perpendicular; the product of slope is - 1; 4. Because m is on the straight line OQ, so y / x = Y0 / x0

Given the fixed point P (2,0), the moving point q is on the circle x ^ 2 + y ^ 2 = 9, and the vertical bisector of PQ intersects OQ at the point m, find the trajectory of the moving point m?

I'll tell you the idea: transfer method
1. Let m (x, y) Q (x0, Y0); the midpoint of PQ be n;
2. First write KPQ, then knm;
3. Because the two lines are perpendicular, the product of the slopes is - 1;
4. Because m is on the line OQ, Y / x = Y0 / x0;
5. The simultaneous solution of x0, Y0;
6. Take Q into the equation of the element

Given that the point P (x0, Y0) is a point on the ellipse x28 + y24 = 1, and the coordinate of point a is (6, 0), the trajectory equation of point m in line PA is obtained

Let m (x, y) be the midpoint of the line PA, then x = x0 + 62y = Y0 + 02, and the solution is x0 = 2x-6y0 = 2Y. ∵ point P (x0, Y0) is a point on the ellipse x28 + y24 = 1, ∵ X208 + y204 = 1. Substituting x0 = 2x-6y0 = 2Y into the above equation, we can get (2x-6) 28 + (2Y) 24 = 1, which is (x-3) 22 + y2 = 1

Given that the ellipse x ^ 2 / 4 + y ^ 2 = 1, P is a moving point on the ellipse, and the coordinate of point a is (1,1 / 2), then the trajectory equation of point m in line PA is obtained

Let m (x, y), P (x ', y'), then (1 + X ') / 2 = x, (1 / 2 + y') / 2 = y, so x '= 2x-1, y' = 2y-1 / 2, substitute them into the elliptic equation to get [(2x-1) ^ 2] / 4 + (2y-1 / 2) ^ 2 = 1, or the ellipse is only centered at (1 / 2,1 / 4), not at the origin

Given that the focus of the ellipse is F1 (- √ 3,0) m, and the right vertex is p (2,0), set point a (1,0.5). If P is a moving point on the ellipse, find the trajectory equation of point m in line PA

In the ellipse: C = √ 3, a = 2 = = > b = 1, let P (2cos α, sin α) Elliptic parameter equation m (COS α + 0.5, 0.5sin α + 0.25) = = > that is: x = cos α + 0.5, y = 0.5sin α + 0.25 = = > cos α = x-0.5, sin α = 2y-0.5 = = > (COS α) ^ 2 + (sin α) ^ 2 = (x-0.5) ^ 2 + (2y-0.5) ^ 2 = 1

The midpoint of line AB is m AB = 6 PA + Pb = 8
Finding the maximum and minimum of PM~

Take M point as the origin, AB as the x-axis, a and B as the focus to establish the coordinate system
AB=2C=6
PA+PB=2a=8
The solution is a = 4, C = 3
a^2=b^2+c^2
Then B = √ (a ^ 2-C ^ 2) = √ 7
The elliptic equation is
x^2/16+y^2/7=1
Pmmin is the length of short axis, B = √ 7
Pmmax is the length of the major axis, a = 4