It is known that the center of the ellipse e is at the origin, the focus is on the x-axis, the minimum distance between the point on the ellipse and the focus is 1, and the eccentricity e = 1 / 2 The line L: y = K (x-1) (k is not 0) intersects the ellipse e at different points P and Q (1) Solving elliptic e-equation (2) Find the range of intercept of vertical bisector of line PQ on y-axis (3) Question: is there a fixed point m on the x-axis so that the vector MP * MQ is a fixed value? If so, find out the coordinates of the fixed point m; if not, explain the reason (1) x ^ 2 / 4 + y ^ 3 = 1 has been obtained Need (2); (3) detailed explanation

2) When y = K (x-1) and x ^ 2 / 4 + y ^ 3 = 1, we get (4K ^ 2 + 3) x ^ 2-8k ^ 2x + 4K ^ 2-12 = 0. According to WIDA's theorem, when X1 + x2 = 8K ^ 2 / (4K ^ 2 + 3) ∧ PQ midpoint is (4K ^ 2 / (4K ^ 2 + 3), - 3K / (4K ^ 2 + 3)) ∧ PQ vertical bisector is y = (- 1 / k) (x-4k ^ 2 / (4K ^ 2 + 3)) - 3K / (4K ^ 2 + 3) x = 0, y = K / (4K ^ 2 + 3) ∧
It is known that the center of the ellipse C is at the origin, the focus is on the x-axis, the minimum distance from the point on C to the left focus is 1, and the eccentricity is e, which is 1 / 2
If the radius of left focus = a + ex, the point on C with the smallest distance from the left focus to the left focus is the left vertex ﹣ a-c = 1, e = C / a = 0.5 ﹣ a = 2, C = 1, B ^ 2 = a ^ 2-C ^ 2 = 4-1 = 3 ﹣ the equation of C is (x ^ 2) / 4 + (y ^ 2) / 3 = 1
The elliptic standard equation with the same focus as the ellipse x ^ 2 / 2 + y ^ 2 = 1 and passing through the point (1,3 / 2)
Given that the two vertices of an ellipse are (- 5,0) (5,0) and one focus is (3,0), find his standard equation
From focus
C=3
The focus is on the x-axis
Then a = 5
a²=25
b²=a²-c²=16
So x & sup2 / 25 + Y & sup2 / 16 = 1
B=4
Equation 25 / A2 + 16 / B2 = 1
It is known that the intersection of the line L1: y = KX + K + 2 and the line L2: y = - 2x + 4 is in the first quadrant. How can we get the range slope of the real number k
By solving the equations of line L1 and line L2, we get: x = (2-k) / (K + 2), because the intersection point is in the first quadrant, we know x > 0, so we get: (2-k) / (K + 2) > 0, that is, (K-2) / (K + 2) < 0. It is equivalent to (K-2) (K + 2) < 0. To solve this inequality, we get: - 2 < K < 2
This is the origin of the range of slope K
It is known that sina + cosa = 7 / 13 and Wu / 2
(1) let Sina cosa = X. be combined with sina + cosa = 7 / 13, and the solution is Sina = (7 + 13X) / 26, cosa = (7-13x) / 26.; [(7 + 13X) / 26] & sup2; + [(7-13x) / 26] & sup2; = Sin & sup2; a + cos & sup2; a = 1. = = = = > X & sup2; = 289 / 169. = = = > | x | = 17 / 13. ∵ π / 2 < a < π, ∵ Sina > 0. ∵ Sina cosa > 0. ∵ Sina cosa = x = 17 / 13. (2) from the above, we can see that x = 17 / 13, And Tana = Sina / cosa = (7 + 13X) / (7-13x) = (7 + 17) / (7-17) = 24 / (- 10) = - 12 / 5.)
Sina + cosa = 7 / 13 square
1+2sinacosa=(7/13)^2
2sinacosa=-120/169
Wu / 20
Sina cosa = radical (289 / 169) = 17 / 13
Two
Sina + cosa and Sina cosa can be used to find the solutions of sin and COS
Sina + cosa = 7 / 13, the square of both sides will become 1 + 2sinacosa = 49 / 169; calculate 2sinacosa
Sina cosa, if you square it, it will become 1-2 sinacosa, and you can substitute the one above into the root
Second, it's very simple. It's known that sina + cosa and Sina cosa can find the sum of sin and COS
The four vertices of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are a, B, C and D. if the inscribed circle of the quadrilateral ABCD just passes through the focus of the ellipse, the eccentricity of the ellipse is equal to ()
A. 22B. 5+12C. 5−12D. 3−52
According to the meaning of the title, the quadrilateral ABCD is a parallelogram, then the center of its inscribed circle is the coordinate origin; the radius of the inscribed circle of the quadrilateral ABCD is the height of the hypotenuse AB in RT △ AOB, according to the meaning of the title, it is easy to get, Ao = a, OB = B; then r = aba2 + B2; according to the meaning of the title, its inscribed circle just passes through the focus of the ellipse, that is, C = r = aba2 + B2; and from A2 = B2 + C2; simultaneously, we can get: e = CA = 5 − 12; so we choose C
If the intersection of the line y = KX + 2K + 1 and the line y = − 12x + 2 is in the first quadrant, then the value range of K is ()
A. −12＜k＜12B. −16＜k＜12C. k＞12D. k＞−12
The intersection of two straight lines is: y = KX + 2K + 1y = − 12x + 2, the solution of the equations is: x = 2 − 4k2k + 1y = 6K + 12K + 1, ∵ the intersection of the straight line y = KX + 2K + 1 and the straight line y = − 12x + 2 is in the first quadrant, ∵ 2 − 4k2k + 1 ＞ 06k + 12K + 1 ＞ 0, the solution of the inequalities is: − 16 ＜ K ＜ 12, so choose B
Given Sina = 8 / 17, a is the second quadrant angle, find sin (a + Wu / 6)
The second quadrant angle Sina = 8 / 17 cosa = - 15 / 17
The result of sin (a + Pai / 6) = sinacos (PAI / 6) + cosasin (PAI / 6) = (8 radical 3-15) / 34 is positive
If a is the second quadrant angle, then cosa = - 15 / 17, sin (a + Wu / 6) = Sina * cos30 + cosa * sin30 = (8 * sqrt (3) - 15) / 34
Using the expansion formula of sin (x + &): sin (a + Pai / 6) = sinacos (PAI / 6) + cosasin (PAI / 6) = 0.849
If the ellipse takes the diagonal vertex A and C of square ABCD as the focus and passes through the midpoint of each side, calculate the eccentricity
Let the four vertices of the square be F1 (- C, 0), P (0, c), F2 (C, 0), q (0, - C), then the equation of the ellipse is x ^ 2 / A ^ 2 + y ^ 2 / (a ^ 2-C ^ 2) = 1. - > (a ^ 2-C ^ 2) x ^ 2 + A ^ 2 * y ^ 2 = a ^ 2 * (a ^ 2-C ^ 2) the midpoint m (C / 2, C / 2) of f1p is on the ellipse

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