![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
If the moving point P satisfies Po * Po = PA * Pb, the range of PA and Pb can be obtained A. B is the intersection point of a circle and X axis (Note: the origin of the circle is arbitrary), P is a moving point of the circle. The moving point P satisfies Po * Po = PA * Pb?
In the rectangular coordinate system xoy, the circle with o as the center is tangent to the straight line: X - (√ 3) y = 4. The circle O and the X axis intersect at two points a and B, and the inner moving point P of the circle makes | PA |, | Po |, | Pb | into G.P series. The range of vector PA and vector Pb is calculated. The straight line: X - (√ 3) y = 4, the slope is = 1 / √ 3, and the vertical line passing through O is y = - √ 3x, two parallel lines
It is known that the circle ox2 + y2 = 4, the intersection of circle O and X axis A, B. the moving point P in the garden satisfies | Po | 2 = | PA |, # 8226; | Pb|
How to simplify this formula
First, the coordinates of point a are (2,0), and the coordinates of point B are (- 2,0). Let the coordinates of point p be (x, y), then | Po | 2 = x ^ 2 + y ^ 2, and | PA | 2 = x ^ 2 + y ^ 2|
=Sqrt ((X-2) ^ 2 + y ^ 2), | Pb | = sqrt ((x + 2) ^ 2 + y ^ 2). Substitute into the equation | Po | 2 = | PA | & # 8226; | Pb |, square both sides at the same time, and then do a simple operation to get x ^ 2 + y ^ 2 = 2
Sqrt means square root
^It means power operation
It is known that P is the moving point a on the center of the circle C: x ^ 2 + y ^ 2 = 1, B is the intersection of the circle C and the Y axis, and the straight lines PA and Pb intersect the X axis m and N respectively
The intersection line y = - 2 at e, F, whether the circle with EF as diameter passes through the fixed point, if so, the coordinates are obtained
Let C: x ^ 2 + y ^ 2 = 1 intersect Y-axis at a (0,1), B (0, - 1). Let P (COSA, Sina), then the slope of PA = (sina-1) / cosa, PA: y = (sina-1) x / cosa + 1, intersects with the line y = - 2 at e (- 3cosa / (sina-1), - 2). Similarly, Pb: y = (Sina + 1) x / cosa-1, intersects with the line y = - 2 at f (- cosa / (Sina + 1), - 2)
It is known that the line L with direction vector E = (1, √ 3) passes through the focus of a (0, - 2 √ 3) and ellipse C: x ^ / A ^ + y ^ / b ^ = 1 (a ＞ B ＞ 0), and the symmetry point of ellipse C center about line L is on the right directrix of ellipse C
① The equation of finding ellipse
② Whether there is a straight line m intersecting ellipse C passing through point E (- 2,0) at points m and N, satisfying,
Vector om × vector on = (4 √ 6) / 3cot ∠ mon (o is the origin). If it exists, find the equation of straight line m; if it does not exist, explain the reason
① The straight line L is: y + 2 √ 3 = √ 3x, that is: y = √ 3x-2 √ 3
It can only be replaced by the right focus of the ellipse (C, 0): C = 2
The right directrix of the ellipse is x = a ^ 2 / C. let the symmetric point of the center of the ellipse C with respect to the line l be P
The abscissa of P is a ^ 2 / C, and the straight line OP: y = - √ 3 / 3x, so p is (a ^ 2 / C, - √ 3A ^ 2 / (3C))
The midpoint of op (a ^ 2 / 2C, - √ 3A ^ 2 / (6c)) is on L: √ 3A ^ 2 / (6c) = √ 3A ^ 2 / C-2 √ 3
So: A ^ 2 = 6, B ^ 2 = 2
The equation of ellipse is x ^ 2 / 6 + y ^ 2 / 3 = 1;
② Let ∠ mon = θ, vector om · vector on = ι om ι on ι cos θ = 4 √ 6 / 3cot θ
│OM││ON│sinθ=4√6/3=2S△MON
Let the line Mn be: KY = x + 2, and the distance from the origin to Mn be: D = 2 / √ (1 + K ^ 2)
Substituting a straight line into an ellipse: (3 + K ^ 2) y ^ 2-4ky-2 = 0
y1+y2=4k/(3+k^2),y1y2=2/(3+k^2)
│MN│=√[(1+k^2)(y1-y2)^2=√(1+k^2)*√[(y1+y2)^2-4y1y2]
4√6/3=2S△MON=d*│MN│=√[(y1+y2)^2-4y1y2]
32 / 3 = 16K ^ 2 / (3 + K ^ 2) ^ 2-8 / (3 + K ^ 2), this equation has no real root, so the straight line m does not exist
Note: vector has two multiplication forms, one is dot product (·) and the other is cross product (×). Dot product is scalar (quantity), and cross product is vector
The focus of the ellipse x ^ 2 / 4 + y ^ 2 / b ^ 2 = 1 (> 0) is on the x-axis, and the symmetric point of the right vertex about X-Y + 4 = 0 is on the left quasilinear of the ellipse
The right vertex of the ellipse is a (2,0), the left quasilinear is x = - A ^ 2 / C = - 4 / C, let the point on the left quasilinear be B (- 4 / C, t), let the midpoint of AB be C, then C = C ((2-4 / C) / 2, t / 2) a, B is symmetric with respect to the straight line X-Y + 4 = 0, then point C must be on the straight line (2-4 / C) / 2-T / 2 + 4 = 0, t = 10-4 / C, and ab ⊥ straight line X-Y + 4 = 0, ⊥ K (AB) =
As shown in the picture, AB.CD Intersection at O, and ab= CD.AD=CB Prove ob = OD
Certification:
Connect BD
∵AB=CD,AD=BC,BD=BD
∴△ABD≌△CDB
∴∠CDB=∠ABD
∴OB=OD
AB, CD intersection point O, and ab = CD, ad = CB, OB = OD
Because AB and CD intersect at o
So, angle AOD = angle cob (equal to vertex angle)
Because AB = CD, ad = CD
Therefore, the triangle AOD is equal to the triangle cob
So, OD = ob
A. B. C is the three points on the circle O, and ∠ ABC = 120 °, ACB = 45 °, the radius of the circle O is 1, and the lengths of the strings AC and ab are calculated
From the sine theorem
AB/sin∠ACB=AC/sin∠ABC=2R
R is the circumcircle of triangle ABC
Directly get AC = root 3
AB = root 2
In the circle O, AB is the diameter, AC is the chord, and point D is on the chord AC with OD = 5, ∠ ADO = 2 ∠ a = 60 °, then ob is longer
∵∠ADO=2∠A=60°
∴∠A=30°
Ψ△ ADO is a right triangle
∴OB=OA=√3*OD=5√3
As shown in the figure, AB is the diameter of ⊙ o, AC is the chord, OD ⊥ AB intersects AC at point D. if ∠ a = 30 °, OD = 20cm, find the length of CD
Solution (1): ∵ OD ⊥ AB, ∠ a = 30 °, OA = OD ⊥ tan30 ° = 203, ad = 2od = 40. ∵ AB is the diameter of ⊙ o, ∵ AB = 403, and ∵ ACB = 90 °, AC = ab ⊥ cos30 ° = 403 × 32 = 60. ∵ DC = ac-ad = 60-40 = 20 (CM). Solution (2): through point O, make OE ⊥ AC at point E, ∵ OD ⊥ AB at point O, ∵ a = 30 °, ∵ ad = 2od = 40, Ao = OD ⊥ tan30 ° = 203. ≁ AE = Ao · cos30 ° = 203 × 32 = 30. ≁ OE ⊥ AC = 30 Solution (3): ∵ OD ⊥ AB at point O, Ao = Bo, ∵ ad = BD. ∵ AB = a = 30 °, and ∵ AB is ⊙ o diameter, ∵ ACB = 90 °, ∵ ABC = 60 °, ∵ ABC = 60 ° - 30 ° = 30 ° = ∵ A. and ∵ AOD = ≌ C = 90 °, and ≌ AOD ≌ BCD. ≌ DC = od = 20 (CM)
RELATED INFORMATIONS
- 1. Hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) f is the right focus, P is a point on the right branch of the hyperbola, P is above the X axis, and M is a point on the left quasilinear O is the origin of coordinates. OmpF is a parallelogram, | PF | = λ| of | (1) Find the relationship between eccentricity e and λ of hyperbola (2) If | ab | = 12, the hyperbolic equation at this time can be obtained (troublesome steps)
- 2. Ellipse X & # 178 / 9 + Y & # 178 / 5 = 1, the line between a point P on the line x = - 9 / 2 and the left focus f intersects the ellipse at two points a and B respectively, if the vector PA = λ AF, Ellipse X & # 178 / 9 + Y & # 178 / 5 = 1, the line between a point P and the left focus f on the straight line x = - 9 / 2 intersects ellipse at two points a and B respectively. If vector PA = λ, vector AF, vector Pb = μ, vector BF, calculate λ + μ
- 3. Let the moving line l be perpendicular to the x-axis and intersect with the ellipse x2 + 2Y2 = 4 at two points a and B. P is the moving point on L which satisfies the vector PA multiplied by the vector Pb = 1. The trajectory equation of point P is obtained
- 4. Given that point P is a point on line AB, the length ratio of AP to Pb is 2:3. If AP = 4cm, find the length of Pb and ab
- 5. Given that the line segment AB = a, P is a point on AB, and AP = ((radical 5-1) / 2) * a, find the ratio of AP to Pb, AB to AP, and ask whether the four line segments AP, Pb, AB and AP are equal
- 6. Solve the plane equation of the straight line x + 3 / 3 = y + 2 / - 2 = Z / 1 and X + 3 / 3 = y + 4 / - 2 = Z + 1 / 1
- 7. If a line passes through point m (a, 3) and point n (1, 2), the equation of the line is solved
- 8. The equation of a straight line passing through a point (1,0-2), parallel to the plane 3x + 4y-z + 6 = 0 and perpendicular to the straight line (x-3) / 1 = (y + 2) / 4 = Z / 1 is obtained
- 9. Solve the plane equation which passes through the point m (3,1, - 2) and the straight line la-4 / 5 = B + 3 / 2 = C / 1
- 10. For example, there are two points a (1,0) and B (- 1,0) on the plane of the graph, and the known equation of the circle is (x-3) ^ 2 + (y-4) ^ 2 = 2 ^ 2 Find the maximum area of ABP1 with a point p1 on the circle and find out the area
- 11. 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
- 12. Let p be a moving point on hyperbola x24-y2 = 1, o be the origin of coordinates, and m be the midpoint of line OP, then the trajectory equation of point m is______ ．
- 13. F1 and F2 are the two focal points of hyperbola, F1 is the focal point of parabola y ^ 2 = 4x, hyperbola passes a (- 2,0), B (2,0), find the trajectory equation of F2 Note the explanation of the value range of X Don't you know what to do?
- 14. The hyperbola C: 2x ^ - y ^ = 2 is known. Find the trajectory equation of the midpoint Q of the chord AB passing through the point m (2,1)
- 15. Given the hyperbola x ＾ 2-y ＾ 2 / 2 = 1. The straight line passing through point a (2,1) intersects with the hyperbola at P1 and P2, find the trajectory equation of the midpoint P of line p1p2
- 16. The center is at the origin, and a hyperbolic equation with focus (5,0) passing (4,0) is Please write the specific process
- 17. From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn. If the perpendicular foot is n, then the trajectory equation of the midpoint P of the line QN is______ ．
- 18. Let p be the interior point of the directed line p1p2, where P1 (2,3) P2 (8,4) is known and p1p / PP2 = PP2 / p1p2, try to find the coordinates of P Vector is the title, thank you, please write the process
- 19. It is known that P is any point on the circle x square plus y square = 4, passing through the point P to make the x-axis PQ. (1) find the trajectory equation of the midpoint m of the line PQ Problem (2) taking point a (- 2,0) as the right angle vertex, we can find the area of the isosceles right triangle ABC which is connected to the locus of point M
- 20. As shown in the figure, the known point P is a point on the square ABCD, PE ⊥ BC, PF ⊥ CD, and the perpendicular feet are e and F, respectively There is no picture,