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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
X ^ 2 / 2-y ^ 2 = 3 (absolute value of x < 2)
Through the point P (1, - 2), make a straight line intersection ellipse X & # 178; + 2Y & # 178; = 8 at two points a and B, PA × Pb = 2 / 3, and find the inclination angle of the straight line
Let the slope of the line be K, then the line is y = K (x-1) - 2
That is y = kx-k-2
Let a (x1, kx1-k-2) B (X2, kx2-k-2)
PA=(x1-1,kx1-k),PB=(x2-1,kx2-k)
PA×PB=(x1-1)(x2-1)+(kx1-k)(kx2-k)=(1+k²)[x1x2-(x1+x2)+1]=2/3
We get: (1 + K & # 178;) [x1x2 - (x1 + x2) + 1] = 2 / 3 (1)
Simultaneous: X & # 178; + 2Y & # 178; = 8, y = kx-k-2
By eliminating y, we get: (1 + 2K & # 178;) - (4K & # 178; + 8K) x + 2K & # 178; + 8K = 0
So X1 + x2 = (4K & # 178; + 8K) / (1 + 2K & # 178;), x1x2 = (2k & # 178; + 8K) / (1 + 2K & # 178;)
Substituting ①, we get the following results
(1+k²)[(2k²+8k)/(1+2k²)-(4k²+8k)/(1+2k²)+1]=2/3
That is, (1 + K & # 178;) / (1 + 2K & # 178;) = 2 / 3
Then: K & # 178; = 1
The solution is k = 1 or K = - 1
When k = 1, the tilt angle is 45 degrees; when k = - 1, the tilt angle is 135 degrees
A: the inclination angle of this line is 45 ° or 135 °
Suppose that the line L passes through the point P (0,3) and the ellipse X29 + y24 = 1 successively intersect at two points a and B, then the value range of APPB is______ ．
As shown in the figure, A2 = 9 and B2 = 4 can be obtained from the ellipse X29 + y24 = 1, and the solution is b = 2. When PA (PB) is tangent to the ellipse, | AP | Pb | = 1. When points a and B are the endpoints of the minor axis of the ellipse, | AP | Pb | = 3 − 23 + 2 = 15. Since the direction of AP and Pb is opposite, | AP | 1 ≤ APPB ≤ − 15
If the straight line with slope 1 intersects with ellipse x ^ 2 + y ^ 2 / 4 = 1 at two points AB, P is the point on line AB, and AP / Pb = 2, then the trajectory equation of point P is obtained
By substituting y = x + m (1) into x ^ 2 + y ^ 2 / 4 = 1, it is concluded that
5x^2+2mx+m^2-4=0,
△=4m^2-20(m^2-4)=16(5-m^2)
Let a (x1, Y1), B (X2, Y2), P (x, y), then
X1,2 = [- m soil 2 √ (5-m ^ 2)] / 5,
From AP / Pb = 2, (x-x1, Y-Y1) = 2 (x2-x, y2-y),
Ψ x = (x1 + 2x2) / 3 = (- 3 / 5) m soil (2 / 5) √ (5-m ^ 2),
By substituting (1), M = Y-X, into the above formula, it is concluded that
(2x+3y)^2=4[5-(y-x)^2],
That is, 8x ^ 2 + 4xy + 13y ^ 2 = 20
In the pyramid p-abcd, CD / / AB, ad ⊥ AB, ad = DC = 1 / 2Ab, BC ⊥ PC, (1) prove PA ⊥ BC
(2) If M is the midpoint of line Pb, we prove that cm ‖ plane pad
(1) ∵ AC = √ (AD ^ 2 + DC ^ 2) = √ 2 / 2Ab, as CE ⊥ AB, similarly, BC = √ 2 / 2Ab, AC ^ 2 + BC ^ 2 = 1 / 2Ab ^ 2 + 1 / 2Ab ^ 2 = AB ^ 2
∴BC⊥AC
∩ BC ⊥ PC, PC ∩ AC in C
Ψ BC ⊥ plane PAC
∴PA⊥BC
(2) Make the point n of PA and connect MNDC
∵ m and N are PA and Pb, respectively
∴MN‖=1/2AB
∵CD‖=1/2AB
The MNDC is a parallelogram
Ψ cm parallel dn
∵ DN ∈ planar pad, CM &; planar pad
‖ cm ‖ planar pad
As shown in the figure, ABCD is a rectangle, PA ⊥ plane ABCD, e is the midpoint of PD, PA = AB = 3, find the sine value of AE and PBC
It seems that there are no conditions for this problem
Take the midpoint N and m of PC and AB, then: AE / / Mn. Because BC ⊥ plane PAB, so plane PAB ⊥ plane PBC, then through the point m, make MH ⊥ Pb in H, so ∠ MNH is the angle between the line AE and plane PBC
It is known that the bottom surface of p-abcd is rhombic and E is the midpoint of PD
Certificate: link AC, BD to o, link OE
Because ABCD is a diamond, O is the midpoint of DB
Then OE is the median line of triangle DPB
So OE is parallel to Pb
Because OE belongs to plane ace
So Pb is parallel to ace
This kind of problem generally borrows the triangle median line
In the pyramid p-abcd, PD ⊥ plane ABCD, PD = DC = BC = 1, ab = 2, AB / / DC, ∠ BCD = 90 ° and PC ⊥ BC are proved
It is proved that because PD ⊥ plane ABCD, BC is in plane ABCD
So PD ⊥ BC
And ∠ BCD = 90 °, that is BC ⊥ CD
So BC ⊥ plane PCD
Because the PC is in the plane PCD
So PC ⊥ BC
It is known that in rectangle ABCD, ab = 3cm, ad = 9cm, fold the rectangle so that point B coincides with point D, and the crease is EF, then the area of △ Abe is______ A．6cm2B．8cm2 C．10cm2D．12cm2．
Fold the rectangle so that the point B coincides with the point D, ∵ be = ed. ∵ ad = 9cm = AE + de = AE + be. ∵ be = 9-ae, according to the Pythagorean theorem, AB2 + AE2 = be2. The area of AE = 4. ∵ Abe is 3 × 4 ∵ 2 = 6
In rectangular paper ABCD, ad = 9, ab = 3, fold it so that point d coincides with point B. from point C to point G, the crease is EF, and calculate the area of △ bef
Let AE be x AB & # 178; + AE & # 178; = be & # 178; = de & # 178; = (9-x) &# 178; = > 9 + X & # 178; = 81-18x + X & # 178; = > 18x = 72 = > x = 4
∴ 9-x=9-4=5=DE=BF
Ψ△ bef area = BF * AB / 2 = 5 * 3 / 2 = 7.5
RELATED INFORMATIONS
- 1. 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
- 2. 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
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- 9. Mathematical problems of plane rectangular coordinate system under seven "Monster eats peas" is a computer game. The symbols in the picture indicate the locations where the "monster" passes successively. If (1,2) is used to indicate the third location where the "monster" passes according to the route indicated by the arrow in the picture, can you show the other locations where the "monster" passes in the same way?
- 10. 1995-1+2-3+4-5+… +1948-1949=______ ．
- 11. 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
- 12. 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)
- 13. 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?
- 14. 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
- 15. 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______ ．
- 16. 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?
- 17. 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)
- 18. 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
- 19. The center is at the origin, and a hyperbolic equation with focus (5,0) passing (4,0) is Please write the specific process
- 20. 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______ ．