It is known that the right focus of ellipse C is F1 (1,0), M is the upper vertex of ellipse C, and O is the coordinate It is known that the right focus of the ellipse C: a squared x squared + B squared y squared = 1 (a > b > 0) is F1 (1,0), M is the upper vertex of the ellipse C, O is the coordinate origin, and the triangle OMF is an isosceles right triangle. (1) the equation for finding the ellipse C (2) whether there is a straight line L intersecting the ellipse at P and Q, and the point F is the perpendicular center of the triangle PQM, Find the equation of the straight line L; if it doesn't exist, explain the reason

It is known that the right focus of ellipse C is F1 (1,0), M is the upper vertex of ellipse C, and O is the coordinate It is known that the right focus of the ellipse C: a squared x squared + B squared y squared = 1 (a > b > 0) is F1 (1,0), M is the upper vertex of the ellipse C, O is the coordinate origin, and the triangle OMF is an isosceles right triangle. (1) the equation for finding the ellipse C (2) whether there is a straight line L intersecting the ellipse at P and Q, and the point F is the perpendicular center of the triangle PQM, Find the equation of the straight line L; if it doesn't exist, explain the reason

Let L: y = x + m, P (x1, Y1) Q (X2, Y2) simultaneous 3x & # 178; + 4mx + 2m & # 178; - 2 = 0 vector AF · MB = (x1-1, Y1) · (X2, y2-1) = x1x2-x2 + y1y2-y1 = x1x2-x2 + (x1 + m) (x2 + m) - (x1 + m) = 2x1x2 + (m-1) (x1 + x2) + M & # 178; - M = 0, we solve the equation

If cos (α + π / 4) = 4 / 5, then Tan α=

cos(α+π/4）=4/5
Then sin (α + π / 4) = 3 / 5 or sin (α + π / 4) = - 3 / 5
Tan (α + π / 4) = 3 / 4 or tan (α + π / 4) = - 3 / 4
(tanα+tanπ/4)/(1-tanαtanπ/4)=3/4
(tanα+1)/(1-tanα)=3/4
4tanα+4=3-3tanα
tanα=-1/7
or
(tanα+tanπ/4)/(1-tanαtanπ/4)=-3/4
(tanα+1)/(1-tanα)=-3/4
4tanα+4=3tanα-3
tanα=-7

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K à ng K à ng open space ch à I ch à ch à C à (difference) (difference) (uneven)
G ā g á g ǎ (Gaga) (Xing GA) Ji à ng Qi á ng Qi ǎ ng (strong) (reluctant)
Lu ò L à o l à (stiff pillow) (fall) J ì N J ì ng (hard) (stiff)

How to prove logn n + 1 times logn n-1

True number > 0, that is n + 1 > 0, n-1 > 0, so n > 1
So logn x increases monotonically at (0, positive infinity)
When 1

What is the Veda theorem

The content of Vieta's theorem
In the univariate quadratic equation AX ^ 2 + BX + C = 0 (a ≠ 0 and △ = B ^ 2-4ac ≥ 0)
Let two roots be X1 and x2
Then X1 + x2 = - B / A
X1*X2=c/a
Cannot be used for line segments
Judge the root of equation with Veda's theorem
If B ^ 2-4ac > 0, the equation has two unequal real roots
If B ^ 2-4ac = 0, then the equation has two equal real roots
If B ^ 2-4ac

Calculating primary school mathematics skillfully
2001×20001999—1999×20002001

2001×20001999—1999×20002001 =(1999+2)×20001999—1999×(20001999+2)=1999×20001999+2×20001999-1999×20001999-1999×2=2×20001999-2×1999=40000000

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rt

Miete, Sieben, Vier, vielleicht, field, nieee, believe, thief, piece, cookie, achievement, Gie,

A problem of analytic geometry in Senior High School
Find the shortest distance between the point (a, 0) and the point on the ellipse (x ^ 2 / 2) + y ^ 2 = 1
The discussion should be classified

Let a point on the ellipse be (√ 2cos θ, sin θ)
Then the shortest distance between the point (a, 0) and the point on the ellipse (x ^ 2 / 2) + y ^ 2 = 1
D ^ 2 = (a - √ 2cos θ) ^ 2 + (sin θ) ^ 2
d^2=(a-√2cosθ）^2+(sinθ）^2
=a^2-2√2acosθ+2（cosθ）^2+(sinθ）^2
=a^2-2√2acosθ+2（cosθ）^2+1-(cosθ）^2
=（cosθ）^2-2√2acosθ+1+a^2
=（cosθ-√2a）^2+1-a^2
Discussion, when a > √ 2 / 2, obviously
When cos θ = 1, d ^ 2 is the minimum
Then d ^ 2 = (a ^ 2-2 √ 2A + 2)
Continue the discussion,
When a > 2, the minimum value of D is a - √ 2
When √ 2 / 2

3x of 4-2x of 5 = 7 of 8
x－25％x＝48

(3/4-2/5)x=7/8
(7/20)x=7/8
x=7/8÷7/20
x=5/2
（1-25%)x=48
0.75x=48
x=48÷0.75
x=64

The solutions of the equations x = 2Y + 1,2x-y = 11,

x=2y+1
x-2y=1 （1）
,2x-y=11 （2）
(2) * 2 - (1) get
3x=21
x=7
Substituting (1)
y=3