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Given that the parabola x 2 = 4Y, point P is a moving point on the parabola, and the coordinates of point a are (12,6), the minimum value of the sum of the distance from point P to point a and the distance from point P to X axis is obtained
Substituting x = 12 into x2 = 4Y, we get y = 36 ＞ 6, so point a is outside the parabola. The focus of the parabola is f (0,1), and the collimator is l: y = - 1. As shown in the figure, PA + PC = PA + PB-1 = PA + PF-1 when passing through point P and making Pb ⊥ l at point B and intersecting X axis at point C
The minimum sum of the distance from P to a (2,10) and the distance from P to the focus can be obtained by moving P on the parabola y = 2x & # 178
The standard equation of parabola is: X & # 178; = Y / 2, and the Quasilinear is x = - 1 / 8
Finding the minimum value of PA + PF
Let the distance from P to the guide line x = - 1 / 8 be d
Then pf = D
Therefore, the minimum value of PA + D is obtained
It is easy to draw. When p is on the vertical line, PA + D is the minimum
The minimum value is the distance from a to the guide line, which is 2 + 1 / 8 = 17 / 8
Therefore, the minimum value is 17 / 8
If the line y = 2x + K cuts the parabola y ^ 2 = 4x, the chord is ab, and the trajectory equation of the midpoint m of the chord is obtained
Substituting
(2x+k)^2=4x
4x^2+4kx+k^2=4x
4x^2+(4k-4)x+k^2=0
x1+x2=-(4k-4)/4=1-k
y=2x+k
So Y1 + y2 = 2x1 + K + 2x2 + k = 2 (x1 + x2) + 2K = 2-2k + 2K = 2
M is the midpoint, so x = (x1 + x2) / 2, y = (Y1 + Y2) / 2
So x = (1-k) / 2, y = 1
Y = 2x + K, so k = y-2x
So x = (1-y + 2x) / 2
We also get y = 1
A straight line and a parabola have intersections
So 4x ^ 2 + (4k-4) x + K ^ 2 = 0 has a solution
So 16K ^ 2-32k + 16-16k ^ 2 > = 0
k
If the distance difference between the moving point P and the point F1 (0, - 3) F2 (0,3) is 1, then the trajectory equation of the moving point P is
According to the geometric definition of hyperbola, the trajectory of P is hyperbola
Let x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
C = 3, | pf1-pf2 | = 2A = 1, a ^ 2 = 1 / 4
Then B ^ 2 = C ^ 2-A ^ 2 = 35 / 4
So the trajectory equation of the moving point P is x ^ 2 / (1 / 4) - y ^ 2 / (35 / 4) = 1
RELATED INFORMATIONS
- 1. Given that the distance difference between the moving point P and the points F1 (- 5,0) and F2 (5,0) is 6, then the trajectory equation of point P is?
- 2. Given two fixed points F1 (- 5,0), F2 (5,0), find the trajectory equation of point P whose absolute value of the difference between the distances of F1 and F2 is 6
- 3. The trajectory equation of the point on the plane where the absolute value of the distance difference between two fixed points F1 (- 7,0) F2 (7,0) is equal to 10 2c=14 c=7 c^2=49 2a=10 a=5 a^2=25 b^2=49-25=24 x^2/25+y^2/24=1 But the answer is x ^ 2 / 25-y ^ 2 / 24 = 1 Shouldn't the + sign be used in the middle? Is the trajectory equation + or -?
- 4. Let F1 and F2 be fixed points, | F1F2 | = 6, and the moving point m satisfy | MF1 | + | MF2 | = 6, then the trajectory of the moving point m is () A. Ellipse B. straight line C. circle D. line segment
- 5. There are two fixed points F1 (0, - 4) (0, - 4) in the plane, and the moving point satisfies the trajectory equation of mf1-mf2 absolute value = 8, M
- 6. It is known that the absolute value of the difference between two points F 1 (- 5,0) f 2 (5,0) and them is the moving point trajectory equation of G
- 7. It is known that F 1 and F 2 are the left and right focal points of the hyperbola X & # 178 / A & # 178; - Y & # 178 / B & # 178; = 1, and P (x, y) is a moving point on the right branch of the hyperbola, The minimum value of Pf1 is 8. The minimum value of vector Pf1 and vector PF2 is - 16. Solve hyperbolic equation
- 8. Given that the sum of the distances between the moving point P and the two foci F1 and F2 of hyperbola (x) 2 / 2 - (y) 2 / 3 = 1 is 6, the equation for finding the trajectory C of point P is obtained. If D (0,3), m and N are known to be on C and the vector DM = $vector DM, the value range of $is obtained Where did you copy it? It's totally wrong
- 9. Given that the moving point P of F1 (- radical 3,0) F2 (radical 3,0) satisfies | Pf1 | + | PF2 | = 4, find the maximum and minimum of vector Pf1 * vector PF2
- 10. As shown in the figure, it is known that in the plane rectangular coordinate system, point a (0,2) and point B (2,0) pass through the intersection line ab of the origin at point C, Let BC = t, the coordinates of point P (2, y) (1) find the coordinates of point C (expressed by the algebraic formula containing T) (2) find the function analytic formula of Y with respect to t, and write out the value range of T. (3) when the triangle PBC is an isosceles triangle, find the coordinates of point P
- 11. If the sum of the distances from a point a to B (3,2) on the parabola y2 = 4x to the focus is the smallest, then the coordinate of point a is______ ．
- 12. The coordinates of the point on the parabola y2 = 4x with the closest distance to its focus are
- 13. Find a point P on the ellipse x ^ 2 / 20 + y ^ 2 / 56 = 1, so that the line between point P and two focal points is perpendicular to each other
- 14. The ellipse with focus on X axis X ^ 2 / 4A + y ^ 2 / (a ^ 2 + 1) = 1, and the maximum eccentricity is
- 15. The focus of ellipse X / 5A + Y / (4a + 1) = 1 is on the x-axis, and the value range of eccentricity is calculated
- 16. It is known that the equation of ellipse C is x ^ 2 + 2 (y ^ 2) = 1 Let the left and right focal points of ellipse C be F1 and F2 respectively, and the line L passing through point F1 intersects the ellipse at points m and N. take F2m and f2n as the adjacent sides to make the parallelogram mf2np, and find the maximum length of the diagonal F2P of the parallelogram
- 17. Given the eccentricity of ellipse C E = √ 6 / 3, a Quasilinear equation is x = 3 √ 2 / 2 Let P satisfy: OP vector = om vector + on vector, where m and N are the points on the ellipse, and the product of the slopes of OM and on is - 1 / 3. Question: are there two fixed points a and B, so that PA + Pb is the fixed value? If so, find the coordinates of a and B; if not, explain the reason
- 18. It is known that the center of ellipse C is the origin of coordinate, the length of minor axis is 2, and a Quasilinear equation is l: x = 2
- 19. If a line passing through a focus F1 of ellipse 4x2 + 2Y2 = 1 intersects with ellipse at two points a and B, then a and B and another focus F2 of ellipse form △ abf2, then the perimeter of △ abf2 is () A. 2B. 22C. 2D. 1
- 20. The focus coordinates of the ellipse X25 + y24 = 1 are______ ．