It is proved that the volume bounded by the tangent plane of any point on the cubic (a > o) of the surface XYZ = A and the three coordinate planes is a certain number The answer is: surface XYZ = A & # 179; the normal direction of (x0, Y0, Z0) is {y0z0, z0x0, x0y0} The tangent plane is: y0z0 (x-x0) + z0x0 (y-y0) + x0y0 (z-z0) = 0 Its intercept on three coordinate axes are 3x0, 3y0 and 3z0 The volume of the tetrahedron bounded by the tangent plane and three coordinate planes is: 27x0y0z0 / 6 = 9A & # / 2 My question is, how do you calculate the intercept above!

The intercept in the answer is set
It's like Veda's theorem
Just substitute x = 0, y = 0

It is proved that the volume bounded by the tangent plane of any point on the curve XYZ = 1 and three coordinate planes is constant?

Let the coordinates of any point on the surface (x0, Y0, Z0)
Satisfy x0 * Y0 * Z0 = 1
The normal vector at this point = (Y0 * Z0, x0 * Z0, x0 * Y0)
The tangent plane equation is as follows
y0*z0*(x-x0)+x0*z0*(y-y0)+x0*y0*(z-z0)=0
The plane intersects the X, y and Z axes to form a tetrahedron
Substitute x0 = 0, Y0 = 0 to get z = 3 * Z0
Similarly, we can get: x = 3 * x0, y = 3 * Y0
The lengths of three perpendicular edges of the tetrahedron are l (x) = 3 * x0, l (y) = 3 * Y0, l (z) = 3 * Z0
Ψ volume
=S(xy)*l(z)/3
=l(x)*l(y)*l(z)/6
=(3^3)*(x0*y0*z0)/6
=9
The volume enclosed by three coordinate planes is a certain number 9

XYZ = XY + 9, XY + 9 = XZ + 15?

1. Xyz-xy = 9, that is, 2xyz-2xy = 18,
2. Xy-xz = 6, that is, 3xy-3xz = 18,
3. Let (1) - (2), we get: 3xyz-2xy-3xy + 3xz = 0, that is, X (3yz-5y + 3Z) = 0, we get x = 0, so XYZ = 0

XYZ yzx = XY find the value of XYZ

yzx
+ xy
-------
xyz
Suppose x + Y10 x + y = Z + 10 Z + X + 1 = y + 10 x = y + 1 has decimal rounding

ABC is a triangle. AE = 2 / 3AB, BD = DC, what is the area ratio of triangle bed and quadrilateral ACDE?

When passing point a, am is perpendicular to BC, and when passing point E, en is perpendicular to BC
EN is parallel to am, AE = 2 / 3AB, so en = 1 / 3aD
BD = DC, so BD = 1 / 2BC,
S1:S2=1/2EN*BD:1/2AD*BC=1:6
So S1: S3 = 1: (6-1) = 1:5

A parallelogram, it contains three triangles, of which the area of two blank triangles are 16 and 24 respectively, what is the area of the middle shadow triangle

Hello, gwa333222111
The middle shadow triangle has the same base and height as the parallelogram, so the area is half of the parallelogram
Then the area of the blank triangle is the other half of the area of the parallelogram, and the sum of their areas is equal to the area of the shadow triangle
The area of the middle shadow triangle is: 16 + 24 = 40

A quadrilateral, with diagonal lines, forms four triangles. The areas of the two pairs of top triangles are 4 and 9 respectively. What is the minimum area of the quadrilateral

Let a diagonal line AC be divided into two segments of length m and N by another diagonal line BD. the distance between B and AC is B, and the distance between D and AC is d
bm/2=4,dn/2=9
The quadrilateral area method: S = D (M + n) / 2 + B (M + n) / 2 = (B + D) (M + n) / 2
Using mean inequality (I don't know if you've learned it or not)
S ≥ 2 (radical bdmn) = 24
So the minimum area of a quadrilateral is 24

The quadrilateral is divided into four triangles by two diagonals. The areas of three triangles are 25, 20 and 30 respectively. The shadow is calculated

30×（20÷25）＝75/2＝37.5

The area of the parallelogram in the figure below is 72 square decimeters. What is the area of the triangle in the shadow?

In the figure, the area of the shadow part is 15 square decimeters. How many square decimeters is the area of the blank triangle? How many square decimeters is the area of the parallelogram?
fast

What about the picture?