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

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


Suppose P is on the x-axis with a special value



If the line L passes through the fixed point P (2,1), if the line L respectively intersects the X axis and the positive half axis of the Y axis intersects the two points AB, when the vector PA * and the vector Pb are the maximum, the equation of the line L is obtained


Let the line Y-1 = K (X-2) and use the image K



P is a point outside the plane of square ABCD, PA = Pb = PC = PD = AB, m, n are points on PA and BD respectively, and PN / MA = 1 / 3 (use vector to answer questions!)
(I) to prove: Mn / / plane PBC; (II) to find the angle between Mn and AD


And the key condition of PN / MA = 1 / 3 has leakage error



The line L passing through the point P (1, - 2) with an inclination angle of 45 ° intersects the ellipse x ^ 2 + 2Y ^ 2 = 8 at two points a and B, and finds the value of | PA | multiplied by | Pb |
The topic is in the parametric equation paper


Let the linear equation be: x = 2 & frac12 / 2T + 1
y=2½/2t-2
Then we take it into the elliptic equation and get the | T1 * T2 | = | PA | * | Pb|
The geometric meaning of the linear parametric equation is the distance to the fixed point (the parameter t is that distance). If t is positive above the fixed point, it is negative below the fixed point,



As shown in the figure, ∠ AOC = 140 ° od bisects ∠ AOC, ∠ AOB is a right angle, find the degree of BOD
(not shown)


Because ∠ AOC = 140 ° OC bisects ∠ AOC
Therefore, AOD = 70 degree
And because ∠ AOB = 90 degree
So ∠ BOD = ∠ AOB - ∠ AOD = 90 ° - 70 ° = 20 °



It is known that: as shown in Fig. 1, ∠ AOB = 70 ° (1) as shown in Fig. 2, OC is a ray in ∠ AOB, OD bisects ∠ AOC, if ∠ BOD = 40 °, calculate the degree of ∠ BOC;
(2) If ∠ BOD = ∠ BOC (∠ BOC < 45 °) and ∠ AOD = half ∠ AOC, please draw a graph and find the degree of ∠ BOC


(1) ∵ - AOB = 70 °, ∵ - BOD = 40 °, ∵ - AOD = ∵ AOB - ∵ BOD = 70 ° - 40 ° = 30 °, ∵ od is the bisector of ∵ AOC, ∵ AOC = 2 ∵ AOD = 60 °,