Can we prove that the line m is perpendicular to the line n?

I think what you said should be that the line m is parallel to the plane α, and the line n is perpendicular to the plane α. This can prove that the straight line m is perpendicular to the straight line n. The reasons are as follows: first, you can find a straight line C in the plane a (assuming) is parallel to the line M. then, because the line n is perpendicular to the plane a, the straight line n is perpendicular to the line C, and because the line m is parallel to the line C, the line m is perpendicular to the line n

High school mathematics analytic geometry The focus of hyperbola C is (plus or minus 3,0), passing through the point P on the straight line x + Y-1 = 0, find out the equation of hyperbola C with the longest real axis~

Let the left focus be F1, the right focus is F2, the symmetric point about the straight line x + Y-1 = 0 is m, connecting f1m, the intersection point with the line x + Y-1 = 0 is the point P, and the value of pf1-pm is 2a, and C ^ 2 = 3 is known, so B ^ 2 can be obtained correspondingly, and the hyperbolic equation is obtained

It is known that the line L1: x + Y-A = 0 L2: X-ay + 2A = 0 (a > 0) intersects point P, line L1 intersects X axis at point B, line L2 intersects X axis and Y axis at point a and point C respectively (1) If the area of the triangle PAB is > = 6, find the value range of A (2) Under the condition of (1), when what is the value of a, the circumscribed circle area of triangle ABC is the smallest? And find the equation of this circle

1) By drawing the graph, we can get AB = 3A (a > 0) and X + Y-A = 0; X-ay + 2A = 0, and y = 3A / (1 + a), that is, point P
The area of the triangle PAB on AB is 3A / (1 + a) multiplied by 3a
Solving inequality 3A / (1 + a) * 3A > = 6 and a > 0, so a > = (1 + √ 7) / 3
2) It can be calculated that the distance from C to AB is 2. With the increase of AB, the area of circle passing through ABP three points increases when a is the smallest
The minimum circle equation of circumcircle area of triangle ABC (x + (1 + √ 7) / 6) ^ 2 + (Y - (5 - √ 7) / 9) ^ 2=
(388+61√7)/162

Analytic geometry What is the path length of the ellipse hyperbola parabola? Is the path of the string that passes through the focus the shortest? Can you directly use it as the shortest condition when doing a big problem? Well, it must be added because it is on the mobile phone, so at most 20, I am waiting for quality and speed at the same time You're not right on the first floor

1. The path length of ellipse and hyperbola is
|AB|=2b^2/a
(where a is 1 / 2 of the major axis or real axis, and B is 1 / 2 of the minor axis or imaginary axis. This conclusion can be reached whether the focus of an ellipse or hyperbola is on the x-axis or the y-axis.)
2. The path length of parabola is
|AB|=4p
Where the focal distance is 1 / 2
3. The diameter of the chord passing through the focus is the shortest
This conclusion is only applicable to ellipse and parabola, but also to hyperbola
If the eccentricity of hyperbola E > root 2, the real axis of the chord passing through the focus is the shortest, that is, the shortest focus chord is 2A
If the eccentricity e of the hyperbola is root 2, then the path is equal to the real axis, and they are the shortest focus chord
If the eccentricity of hyperbola 0A > 0,
|MN|=2ab^2(k^2+1)/[(bk)^2+a^2]
When k = 0, the maximum value of | Mn | is 2A
Let | ab | be the path, then | ab | ≤ Mn | ≤ 2a in the ellipse
If | Mn|

Let o be the coordinate origin, and there are two points P and Q on the circle C: x2 + Y2 + 2x-6y + 1 = 0, which are symmetric about the straight line x + my + 4 = 0 and satisfy op ⊥ OQ, (1) Find the value of M; (2) Find the equation of straight line PQ Why do I want to put the equation of circle into the equation 1 2. Why "Y1 · y2 = B ^ 2-B (x1 + x2) + x1 · x2"

The purpose of substituting the linear equation into the circle equation is to find the intersection point of the straight line and the circle, that is, P, Q, △ 0, to ensure that there are two different intersections
2 Y1 · y2 = B ^ 2-B (x1 + x2) + x1 · X2 is because y = - x + B has been calculated, so Y1 · y2 = (- X1 + b) (- x2 + b) will bring x1x2 in

What is the function of solid geometry?

I think the question is like this, you know, there is plane geometry on the plane, there is a plane rectangular coordinate system, the corresponding function image can be made in the plane rectangular coordinate system

In the cuboid abcd-a1b1c1d1, if the bottom is a square with side length of 2 and height of 4, then the distance from point A1 to section ab1d1 is () A. 8 Three B. 3 Eight C. 4 Three D. 3 Four

As shown in the figure, let a1c1 ∩ b1d1 = O1, ∵ b1d1 ⊥ a1o1, b1d1 ⊥ Aa1, ∩ b1d1 ⊥ plane aa1o1,
Therefore, the plane aa1o1 ⊥ surface ab1d1, the intersection line is AO1, and b1H ⊥ AO1 in H is made through B1 in plane aa1o1,
It is easy to know that the length of a1h is the distance from point A1 to section ab1d1. In RT △ a1o1a, a1o1=
2，
AO1=3
From a1o1 · A1A = h · AO1, a1h = 4
3，
Therefore, C

Find the derivative of the function y = 3 √ (1 + cos2x)

-(2/3)*sin(2*x)/(1+cos(2*x))^(2/3)

A detailed explanation of the derivation of the function y = (1 + cos2x) ^ 3 This is the online answer, just tell me how to get out of the first step "This step is to take 1 + cos2x as a global variable U, i.e., take the derivative of y = u ^ 3." If it is said above, it should be y '= 3U ^ 2, why multiply (1 + cos2x)? Y '= 3 (1 + cos2x) ^ 2 * (1 + cos2x)' (this step) =3(1+cos2X)^2cos2X' =3(1+cos2X)^2(-sin2X)2X' =6(1+cos2X)^2(-sin2X)

Let y = f (U) and u = V (x) have a nonempty domain intersection in which they are continuous and differentiable. Then: the derivative of a function of the form: y = f '(U) * V' (x) = f '[V (x)] * V' (x) = f '[V (x)] * V' (x)

The derivative of the function y = sin2x / 1-cos2x is

y=sin2x/(1-cos2x)
=2sinxcos/(2sin²x)
=cosx/sinx
y'=[(cosx)'*sinx-(sinx)'cosx]/(sin²x)
=(-sinxsinx-cosxcosx)/sin²x
=-1/sin²x
=-csc²x

Can we prove that the line m is perpendicular to the line n?