If θ is an internal angle of a triangle and sin θ + cos θ = 15, then the curve represented by the equation x2sin θ + y2cos θ = 1 is () 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, so, θ ∈ (π 2, π), and | sin θ| > cos θ|, so θ ∈ (π 2, 3, π 4), so cos θ < 0, so x2sin θ + y2cos θ = 1 represents the hyperbola with focus on the x-axis

Let θ∈ (π / 2,5 π / 6), then the curve represented by the equation x ^ 2 / (COS θ - 2) + y ^ 2 / (2-sin θ) = 1 is?
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 the range of this angle belongs to the second quadrant, the cosine of this angle is a negative number less than 0, and then subtracting 2 is of course a negative number. And the sine of this angle is less than a positive number of 1, so subtracting a positive number less than 1 from 2 is certainly not a negative number, and it must be a positive number. In this way, the first term on the left of the equation is a negative number, and the second term is a positive number, Isn't the curve represented by the equation a hyperbola? And the focus is on the Y axis. So the answer is d

Let θ Σ (3 / 4 ππ) be the curve represented by the equation x ^ 2 / sin θ - y ^ 2 / cos θ with respect to XY
Why an ellipse instead of a curve
By the way, what about the ellipse with the focus on the y-axis

When θ ∈ (3 / 4 π, π), sin θ 0, so the essence of this equation is ellipse. When θ ∈ (3 / 4 π, π), using the image of sine function and cosine function, we can compare the size of denominator, or put forward the root sign 2 to simplify sin θ - (- cos θ), which is equal to the root sign 2 times sin (θ + π / 4)

Circle C: x2 + y2-2x-2y-7 = 0, let p be the midpoint of the chord of the circle passing through point (3,3), then the trajectory equation of the moving point P is______ ．

∵ circle C: x2 + y2-2x-2y-7 = 0, the standard equation is (x-1) 2 + (Y-1) 2 = 9, the center of the circle is C (1,1), and the radius r = 3. Let a (3,3), connect PC ∵ p to the midpoint of the chord of the circle passing through point (3,3), and ∵ PC ⊥ AP, then point P can move on the circle with AC as the diameter

If the circle (x-a) 2 + (y-b) 2 = 6 always bisects the circumference of X + y + 2x + 2y-3 = 0, then the trajectory equation of the moving point m (a, b) is ()
Steps to solve the problem
If the circle (x-a) 2 + (y-b) 2 = 6 always bisects the circumference of x2 + Y2 + 2x + 2y-3 = 0, then the trajectory equation of the moving point m (a, b) is
Note: (x-a) 2 -- the square of (x-a), (y-b) 2 -- the square of (y-b), X2 -- the square of X, Y2 -- the square of Y

"X + y + 2x + 2y-3 = 0", are you wrong?

The answer to the trajectory equation of a circle tangent to the line y = 0, X2 + y2-2y = 0 is x2 = 4Y,

Circle F: x2 + y2-2y = 0, that is, X & # 178; + (Y-1) &# 178; = 1
Let m (x, y) be the center of the moving circle and R be the radius
The circle m is tangent to the line y = 0 (x axis)
∴r=|y|
∵ circle m is tangent to Circle F: X & # 178; + (Y-1) &# 178; = 1
| MF | = 1 + r = 1 + | y | (circumscribed)
|MF | = | 1-r | = | 1 - | y | (inscribed, Y > 0)
|MF|=1+|y|==> x²+(y-1)²=(1+|y|)²
∴x²+y²-2y+1=1+2|y|+y²
Ψ X & # 178; = 4Y (Y ≥ 0) or x = 0 (Y ≤ 0)
|MF|=|1-|y| ==>x²+(y-1)²=(1-|y|)²
∴x²+y²-2y+1=1-2|y|+y²
∴x=0(y>0)
The new trajectory equation of moving circle is
X & # 178; = 4Y (parabola) and x = 0 (Y-axis)
Your answer is part of it

The equation x2 + Y2 + 2aX by + C = 0 indicates that the center of the circle is C (2,2) and the radius is 2, then the values of a, B and C are ()
A. 2，4，4B. -2，4，4C. 2，-4，4D. 2，-4，-4

From x2 + Y2 + 2aX by + C = 0, the center coordinate is (- A, B2), and the radius is R2 = B24 + A2 − C. because the center of the circle is C (2, 2), and the radius is 2, the solution is a = - 2, B = 4, C = 4, so B is chosen

1： If (x2 + Y2) 2-5 (x2 + Y2) - 6 = 0, then x2 + y2 =. 2: if the lengths of three sides of a triangle satisfy the equation x2-6x + 8 = 0,
1： If (x2 + Y2) 2-5 (x2 + Y2) - 6 = 0, then x2 + y2 =
2： If the lengths of three sides of a triangle satisfy the equation x2-6x + 8 = 0, then the perimeter of the triangle is
3： We know the equation x2 + (4K + 1) x + 2k-1 = 0,
(1) It is proved that the equation must have two unequal real roots
(2) Let x 1 and x 2 be the two real roots of the equation, and (x 1-2) (x 2-2) = 2k-3, find the value of K

1： Let x2 + y2 = a, then the formula is a square - 5a-6 = 0, a = 6 or = - 1, because x2 + Y2 is greater than or equal to 0, so it is equal to 62: x = 2 or 4, because the sum of the two sides of the triangle must be greater than the third side, so the third side is equal to 4, and the perimeter is 103 (1): because there are two unequal real roots, so square-4ac must be greater than 0square-4ac =

Is the electron cloud model the same as the atomic planet model? Is the proposer the same?

The electron cloud model is different from the atomic planet model: the planet has a fixed orbit; the electron cloud is not a fixed orbit, but the probability of the electron
Atomic planet model proposer: the atomic model of primitive planetary structure was proposed by the French physicist Perrin; the solar system model was proposed by the British physicist Ernest Rutherford
In 1926, Austrian scholar Schrodinger put forward the famous Schrodinger equation of second-order partial differential on the basis of debroy relation. The solution of this equation, if expressed graphically in three-dimensional coordinates, is the electron cloud

If strictly speaking, is the electronic cloud a model? Why?

The electronic cloud is a model
The electron cloud model is the famous Schrodinger equation of second-order partial differential, which was put forward by Austrian scholar Schrodinger in 1926 on the basis of debroy relation. The solution of this equation, if expressed graphically in three-dimensional coordinates, is the electron cloud model
Please refer to inorganic chemistry, edited by Department of inorganic chemistry, Beijing Normal University, 1981 edition, published by people's education press

If θ is an internal angle of a triangle and sin θ + cos θ = 15, then the curve represented by the equation x2sin θ + y2cos θ = 1 is () 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