It is known that M is the equal proportion median of two positive numbers 2 and 8, then the eccentricity of conic x ^ 2 + y ^ 2 / M = 1 is As long as the result (two solutions)

It is known that M is the equal proportion median of two positive numbers 2 and 8, then the eccentricity of conic x ^ 2 + y ^ 2 / M = 1 is As long as the result (two solutions)

M is ± 4, centrifugal rate is √ 5,3 / 4

If there is a point a on the image of the inverse scale function y = KX, its abscissa n makes the equation x2-nx + n-1 = 0 have two equal real roots, and the area of the triangle with points a, B (1,0) and C (4,0) as vertices is equal to 6, then the analytic expression of the inverse scale function is___ ．

Because the equation x2-nx + n-1 = 0 has two equal real roots, so △ = 0, that is n2-4 (n-1) = 0, the solution is N1 = N2 = 2. Let the height of the triangle be h, and because AC = 4-1 = 3, the area of the triangle is 6, so 12 × 3H = 6, the solution is h = 4

The intersection coordinates of the image of quadratic function and the line y = 1 are (- 1,1) and (2,1), and it passes through the point (- 3,6). Find the expression of this function

(- 1,1) and (2,1) are symmetric with respect to the line x = 1 / 2, so the symmetry axis of the quadratic function is x = 1 / 2
Let y = a (x-1 / 2) & sup2; + K
Substituting the points (- 1,1) and (- 3,6), we get: 9A / 4 + k = 1, 49A / 4 + k = 6
The solution is a = 1 / 2, k = - 1 / 8
The expression of this function is y = 1 / 2 (x-1 / 2) & sup2; - 1 / 8
That is: y = 1 / 2x & sup2; - 1 / 2x

Let a straight line L pass through the point P (3,4), and its inclination angle is twice the inclination angle of the root sign 3 of the straight line, which is multiplied by X and then - y + root sign 3 = 0

The root sign 3 is followed by multiplication X and then - y + root sign 3 = 0, slope k = root sign 3, k = Tana=
a=60
Linear slope k: k = tan2a = - √ 3
y=-√3x+4+3√3

The equation for finding the line L whose sine value of inclination angle is equal to root 3 / 2 and passes through point (1,0)

Because the sine value of the tilt angle = √ 3 / 2
So the slope of this line is ± √ 3
Let the line be y = KX + B
So k = + √ 3 or - √ 3
A straight line passes through the point (1.0)
When k = √ 3, B = - √ 3
When k = - √ 3, B = √ 3
So the linear equation is:
Y = {3x -} 3 or y = - {3x +} 3

Let the inclination angle of the straight line l be 60 degrees and pass through the point P (2, 3 under the root sign), then the equation of the straight line L and its intercept on the y-axis can be obtained?

k=tan60=√3
Then y - √ 3 = √ 3 * (X-2)
That is y = √ 3 (x-1)
The intercept on the y-axis is - √ 3

The inclination angle is y = root 2 + 1, twice of the inclination angle, and the intercept on the y-axis is - 2

In line y = √ 2x + 1
k=tana=√2
k'=tan2a=2tana/(1-tana*tana)
=2√2/(1-√2*√2)=-2√2
Because B = - 2
So the linear equation is y = - 2 √ 2x-2

A linear equation with the inclination angle half of the inclination angle of the line y = - root 3 + 1 and the intercept of - 10 on the y-axis

Let the inclination angle be α, and the linear equation be y = KX + B Tan α = - √ 3 α ∈ [0,2 π] - α = 120 ° α / 2 = 60 ° Tan, 60 ° = √ 3  y = √ 3x + B B = - 10  y = √ 3x-10