If the center of the ellipse is the origin, the focus is on the x-axis, the point P is a point on the ellipse, and the projection of P on the x-axis is exactly the origin of the ellipse The center of an ellipse is at the origin, the focus is on the x-axis, P is a point on the ellipse, the projection of P on the x-axis is just the left focus of the ellipse, the line between P and center O is parallel to the line between the right vertex and the upper vertex, and the distance between the left focus and the left vertex is equal to √ 10 - √ 5. Try to find the eccentricity and equation of the ellipse

If the center of the ellipse is the origin, the focus is on the x-axis, the point P is a point on the ellipse, and the projection of P on the x-axis is exactly the origin of the ellipse The center of an ellipse is at the origin, the focus is on the x-axis, P is a point on the ellipse, the projection of P on the x-axis is just the left focus of the ellipse, the line between P and center O is parallel to the line between the right vertex and the upper vertex, and the distance between the left focus and the left vertex is equal to √ 10 - √ 5. Try to find the eccentricity and equation of the ellipse

Elliptic equation x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
P is a point on the ellipse, and the projection of P on the X axis is exactly the left focus of the ellipse,
P(-c,y0),yo

Ellipsoid equation

The equation of quadric surface is not derived (except for several surfaces of revolution, see [note]). In fact, the equation of quadric surface of x0d exists first, and then the graph of it is studied, For the coefficients Q and R of the first-order terms, the same can be done. For the quadratic terms of x0d, there are generally cross terms XY, YZ and ZX. Because they can be eliminated by orthogonal transformation of the quadratic form in linear algebra, they are not discussed in Higher Mathematics at the beginning. Except for the cylindrical equation, the standard form to be discussed is actually only ax ^ 2 + by ^ 2 + CZ ^ 2 + C = 0 and x0d (1) C ≠ 0, It includes ellipsoid, univalent hyperboloid and bilobed hyperboloid; C = 0, including all kinds of cone; ax ^ 2 + by ^ 2 + RZ + C = 0, including elliptic paraboloid and hyperbolic paraboloid (saddle surface is nickname, [note] ① the trajectory of a space point whose sum of distances to two fixed points is a fixed value is a rotating ellipsoid; ② the trajectory of a space point whose difference between distances to two fixed points is a fixed value is a rotating hyperboloid; ③ the trajectory of a space point whose distance to a fixed point is equal to a given plane is a rotating parabolic surface After giving the data, it is easy to deduce the distance formula

The equation of semi rotating ellipsoid is?
There is a semi rotating ellipsoidal pool with a depth of five meters and a diameter of six meters. The pool is full of water with a specific gravity of one. How much work does it need to do to pump all the water in the pool?

First write out the analytic expression of the plane, and then rotate it according to the third axis
（x*x +y*y）/9 + z*z/ 25 =1
I think only half of the ground is cut into two parts. The plane perpendicular to the ground is z-axis, x-axis and y-axis, which coincides with the ground
College Advanced Mathematics Tongji press Volume 2, 5th Edition
List the analytic expression of the ellipse on the xoz plane and rotate it according to the z-axis

Matlab to solve tangent plane and normal equation; draw the ellipsoid and tangent plane and normal graphics
Find the tangent plane and normal equation of the rotating ellipsoid 3x ^ 2 + y ^ 2 + Z ^ 2 = 16 at the point (- 1, - 2,3); draw the graph of the ellipsoid and its tangent plane and normal
In five minutes, it's urgent

SYMS X Y Z; F = 3 * x ^ 2 + y ^ 2 + Z ^ 2-16; NV = Jacobian (F, [x y z]); [x, y, Z] = sphere; mesh (4 / sqrt (3) * x, 4 * y, 4 * z);% ellipse x = - 1; y = - 2; Z = 3; NV = double (subs (NV)); hold onquiver3 (x, y, Z, NV (1), NV (2), NV (3),. 5);% normal vector t = - 1: 5:1; [XX, YY]

Let θ be an internal angle of △ ABC and sin θ + cos θ = 15, then x2sin θ + y2cos θ = 1 denotes ()
A. Ellipse with focus on X-axis B. ellipse with focus on Y-axis C. hyperbola with focus on X-axis D. hyperbola with focus on y-axis

Because θ ∈ (0, π), and sin θ + cos θ = 15, the square of both sides can be obtained, sin θ· cos θ = - 1225 ＜ 0, so θ ∈ (π 2, π), and | sin θ | cos θ |, so sin θ ＞ 0, cos θ ＜ 0, so x2sin θ, + y2cos θ = 1 represents the hyperbola of the focus on the x-axis

Let a be an internal angle of a triangle and Sina + cosa = 1 / 5, then what does the equation x ^ 2sina-y ^ 2cosa = 1 represent

According to the meaning of the question, Sina = 4 / 5, cosa = - 3 / 5
So 4x & # 178 / 5 + 3Y & # 178 / 5 = 1 represents the ellipse with focus on Y axis

High school mathematics knows that a is an internal angle of △ ABC, and Sina + cosa = 1 / 2, then the curve represented by the equation x ^ 2sina-y ^ 2cosa = 1 is

Sina + cosa = 1 / 2. The square of both sides leads to: 1 + 2sinacosa = 1 / 4. = = = > sinacosa = - 3 / 8. ∵ 0 < a < 180 & ordm; ∵ cosa < 0 < Sina. ∵ the curve expressed by the equation is ellipse

The equation x ^ 2 / Sina + y ^ 2 / cosa = 1, a belongs to (0, π), which means ellipse, and the range of finding a

The equation x ^ 2 / Sina + y ^ 2 / cosa = 1, a belongs to (0, π) and represents ellipse
sina>0
cosa>0
A belongs to (0, π)
The range of a: a belongs to (0, π / 2)

(1-cosa) / Sina =? To tan

Sina = 2sina / 2cosa / 2, 1-cosa = 2sina / 2sina / 2, so (1-cosa) / Sina = 2sina / 2sina / 2 / 2sina / 2cosa / 2 = Sina / 2 / cosa / 2 = Tana / 2

If Sina cosa = 1 / 2 Tan a=

Sina cosa = 1 / 2, then Sina = 1 / 2 + cosa is substituted into sin & # 178; a + cos & # 178; a = 1 = then (1 / 2 + COSA) &# 178; + cos & # 178; a = 1, namely 2cos & # 178; a + cosa-3 / 4 = 0, namely 8cos & # 178; a + 4cosa-3 = 0  cosa = (1 + √ 7) / 4 or cosa = (1 - √ 7) / 4 (1) cos = (1 + √ 7) / 4, then Sina = (1 -