How to find the distance from a point P on an ellipse to the left and right guide lines Let's talk about the process

How to find the distance from a point P on an ellipse to the left and right guide lines Let's talk about the process

In this way, you can know the eccentricity of the ellipse. The ratio of the distance from point P to the left focus to the distance from the left quasilinear is equal to the eccentricity
According to the second definition of ellipse, the distance to left focus is the distance from eccentricity * to left guide line, and the distance to right focus is the distance from eccentricity * to right guide line
If the distance between the center and the guide line of the ellipse is equal to 10%, then the standard equation of the ellipse is?
C=6
The focus is on the y-axis
x²/b²+y²/a²=1
Alignment y = ± A & sup2 / C
The center is the origin
So a & sup2 / / C = 10
a²=60
b²=a²-c²=24
So x & sup2 / 24 + Y & sup2 / 60 = 1
The focus of the ellipse is C = 6 on the y-axis
The center is the origin
The distance from the center to the guide line is equal to 10
b^2/c=10
b^2=60
a^2=b^2-c^2=60-36=24
The equation of ellipse
x^2/24+y^2/60=1
Given that the line y = KX is the tangent of y = LNX, then the value of K is
The calculation of derivative~
When the line y = KX is tangent to the curve y = LNX, the slope k is the largest
Let the tangent coordinates be (x0, lnx0)
Then k = y '| x = x0 = 1x0 = lnx0-0x0-0 = lnx0x0 ｛ lnx0 = 1, that is, x0 = e, so the tangent coordinates are (E, 1)
The maximum value of K is 1E
Cot θ / (2cos θ + 1) = 1, find Cos2 θ and Cos2 θ / (1 + sin2 θ)
cosA=-1
cos2A=2(cosA)^2-1=1,
sin2A=0
cos(2A)/(1+sin2A)=1
In the rectangular trapezoid oabc, CB ‖ OA, COA = 90 °, OE = 2EB, CB = 3, OA = 6, Ba = 35, OD = 5. The plane rectangular coordinate system as shown in the figure is established by taking the line of OA and OC as x-axis and y-axis respectively
It is proved that: through point B, BG ⊥ X axis intersects X axis at point G, ∵ CB ∥ OA, ∠ COA = 90 °, CB = 3, ≁ og = 3, ≁ GA = oa-og = 6-3 = 3, BG ⊥ X axis, ≁ in right triangle AGB, BG2 = ab2-ga2 = (35) 2-32 = 36, ≁ BG = 6, then ob = 35 is obtained according to Pythagorean theorem, OE = 2 is obtained from OE = 2be
Given that the line y = KX is the tangent of the curve y = 12x2 + LNX at x = e, then the value of K is ()
A. e+1eB. e−1eC. 2eD. 0
∵ y = 12x2 + LNX, ∵ y ′ = x + 1X, ∵ y ′| x = e = e + 1E. The value of ∵ K is e + 1E
The value of (cot α - 1) / (2cot α + 1) = 1, (Cos2 α) / (1 + sin2 α) =?
(cot α - 1) / (2cot α + 1) = 1cot α - 1 = 2cot α + 1cot α = - 2tan α = - 1 / 2, so (Cos2 α) / (1 + sin2 α) = (1-tan & # 178; α) / (1 + Tan & # 178; α)] / (1 + 2tan α / (1 + Tan & # 178; α)) = (1-tan & # 178; α) / (1 + Tan & # 178; α + 2tan α) = (1-tan α) / (1 + Tan α) = (1 + Tan α) =
Question: in the plane rectangular coordinate system, point O is the origin of coordinates, known as isosceles trapezoid oabc, OA / / BC, point a (4,0), BC = 2, the height of isosceles trapezoid oabc
Q, the line y = -- 1 / 5x + 6 / 5 intersects the line AB at the point P (P, q), and the point m (m, n) is on the line y = -- 1 / 5x + 6 / 5. When n is greater than Q, find the value range of M,
1) According to the conditions, OA is the bottom, BC is the top, OA is on the positive half axis of X, BC is in the first quadrant, if the height is 1, then the ordinate values of B and C are 1. If this trapezoid is isosceles trapezoid, then according to the symmetry characteristics of isosceles trapezoid, we can know that the abscissa of B is 3, and the abscissa of C is 1, so the coordinates of four points of isosceles trapezoid are known, and then the figure is also out
O(0,0) A(4,0) B（3,1） C(1,1)
2) The straight line of AB can be obtained as follows: y = - x + 4
The coordinates of point P are the intersection of (7 / 2,1 / 2) two straight lines, and the equations are solved
Therefore, when n > Q, M
The tangent of y = LNX is y = KX?
y=lnx,y'=1/x
Then the tangent through (a, LNA) is y-lna = K (x-a)
y=kx-ka+lna
So - Ka + LNA = 0
Where k is the function value of Y 'when x = a
So k = 1 / A
So - (1 / a) * a = LNA
lna=1
A=e
So k = 1 / a = 1 / E
Solution;
Let the tangent point be (x0, Y0)
So the tangent point is y = LNX, y = KX
therefore
y0=ln(x0),(1)
y0=kx0,(2)
y=lnx
y'=1/x
k=1/x0,(3)
From (1), (2), (3)
y0=1,x0=e
k=1/e
So k = 1 / E
Sina = Cos2 α, α belongs to (π / 2, π), then the value of cot α is?
sina=cos2α=1-2(sina)^2
Let Sina = t
t=1-2t^2
2t^2+t-1=0
(2t-1)(t+1)=0
t=1/2 or -1
Because α belongs to (π / 2, π)
So Sina = 1 / 2 cosa