Let the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) be F1 and F2 respectively. A is the point on the ellipse, af2 is perpendicular to F1F2, and the distance from the origin o to AF1 is (2) Q1 and Q2 are the two moving points on the ellipse, oq1 is perpendicular to 0q2, the perpendicular od of the straight line q1q2 is made through the origin o, the perpendicular foot is D, and the trajectory of D is obtained

If AF1 ⊥ af2, the eccentricity of the ellipse is?

A is an endpoint of the minor axis of the ellipse, AF1 ⊥ af2
∴∠AF1O=45º , |AF1|=a,|OF1|=c
∴e=c/a=cos45º=√2/2

Point P is a point on the ellipse x2 / 25 + Y2 / 16 = 1, F1 and F2 are the two focal points of the ellipse, and the inscribed circle radius of triangle pf1f2 is 3 / 2

Point P is a point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1, F1 and F2 are the two focal points on the ellipse, and the radius of the inscribed circle of △ pf1f2 is 3 / 2. When point P is above the X axis, what is the ordinate of point p?
x^2/25+y^2/16=1
a^2=25,a=5,
c^2=a^2-b^2=25-16=9,c=3,
Radius of inscribed circle r = 3 / 2
Δ pf1f2 area = [| Pf1 | * r + | PF2 | * r + | F1F2 | * r] * 1 / 2
=r/2*(2a+2c)
=r(a+c)
=3/2*(5+3)
=12
And: △ pf1f2 area = 1 / 2 * | F1F2 | * the ordinate of point P
So, the ordinate of point P = 2 * 12 / 2C = 12 / 3 = 4
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Let the lengths of the three sides of a triangle be a, B, C respectively, and the radius of the inscribed circle be r, then the area of the triangle = (a + B + C) r * 1 / 2

It is known that P is the point on the ellipse x ^ 2 / 16 + y ^ 2 / 12 = 1, and F1 and F2 are the two focal points of the ellipse, if the inscribed circle radius of △ pf1f2 is 1

I don't know what you want to ask. Do as you are asked
F1(-2,0),F2(2,0)
P (x, y) rule
The inscribed circle radius of △ pf1f2 is 1
S△PF1F2=(PF1+PF2+F1F2)r/2
=（8+4)/2=6
S△PF1F2=c*|y|=2|y|=6
==>|y|=3 |x|=2
|PO|=√x^2+y^2=√13

Given that point P is the point in the first quadrant of the ellipse X29 + Y25 = 1, F1 and F2 are the two focal points of the ellipse. If the radius of the inscribed circle of △ pf1f2 is 12, then the coordinate of point P is ()
A. （355，2）B. （3114，54）C. （3598，58）D. （2，54）

Let P (x, y) (x, Y > 0) ∵ △ pf1f2 area s = 12R (| Pf1 | + | PF2 | + | F1F2 |) = 12Y | F1F2 |. 12 (2 × 3 + 2 × 2) = y × 2 × 2, y = 54. Substituting into the elliptic equation, we can get: X29 + (54) 25 = 1, x = 3114. ∵ P (3114, 54)

If the plane of the square ABCD and the plane of the square abef form a dihedral angle of 60 degrees, the cosine of the angle between the line AD and BF is 0

Wrong,
It should be √ 2 / 4

In the regular pyramid p-abcd, the side edge PA = 2Ab is used to find the cosine value of the dihedral angle p-ab-c

Let AB = 2, AP = 4
Take AB as x-axis, ad as y-axis, and the system perpendicular to it is z-axis
AP = (1,1, root 3) AB = (1.0.0)
Let the normal vector E1 of plane PAB = (x, y, z)
Then E1 · AP = 0, E1 · AB = 0
We deduce x + y + root 3 · z = 0, x = 0
We deduce that x = 0, y = - root 3 · Z
Let z = 1, then E1 = (0, - root 3,1)
There is also the normal vector E2 = (0,0,1) of the plane ABCD
e1`e2 1 1
Then cos = ----- = ----- = 0------
le1l le2l 2*1 2
So = Wu / 3

The side length of the diamond ABCD is 10, ∠ ABC = 60 °, and the distance of AC can be obtained by folding the diamond along the diagonal BD into a dihedral angle of 60 °

Let the intersection of diagonals be o
After being folded into dihedral angle, the triangle AOC is an equilateral triangle, so AC = Ao = 10 / 2 = 5

In the diamond ABCD with side length of 1, the angle ABC is 60 ° and the diamond is folded along the diagonal AC so that BD = 1 after folding, what is the volume of the triangular pyramid b-acd?

Because the angle ABC is 60 degrees, the triangle ABC is an equilateral triangle, and AC = AB = BC = 1;
Let two diagonals intersect at point O, the diagonals of the diamond are perpendicular to each other and bisected, and each diagonal bisects a group of diagonals
Od = sin60 * ad = √ 3 / 2
V=1/3*1/2*AC*OD*BD=√3/12

Given the ratio of three sides of triangle ABC is 3:4:5, if the perimeter of similar triangle DEF is 24, then s triangle DEF is 2

∵ the corresponding sides of similar triangles are proportional
The ratio of ∧ DEF is 3:4:5
The perimeter of △ DEF is 24
The three sides are as follows:
24*3/(3+4+5)=6
24*4/(3+4+5)=8
24*5/(3+4+5)=10
∵6^2+8^2=10^2
Ψ Δ DEF is a right triangle
∴S△DEF=1/2*6*8=24