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It is known that the center of ellipse C is at the origin and the focus is on the x-axis. The coordinates of a vertex B of ellipse C are (0,1) and the eccentricity is equal to 22. The line L with slope 1 intersects ellipse C at two points m and n. (1) find the equation of ellipse C; (2) ask whether the right focus F of ellipse C can be the center of gravity of △ BMN? If possible, find out the equation of line L; if not, explain the reason
(1) Let the equation of ellipse C be x2a2 + y2b2 = 1 (a ＞ B ＞ 0), then we know from the meaning that B = 1. That is, 1-1a2 = 22. That is, the equation of ellipse C is (2) suppose that the right focus F of ellipse C can be the center of gravity of △ BMN, let the linear l equation be y = x + m, substitute it into the elliptic equation, eliminate y to get 3x2 + 4mx + 2m2-2 = 0, and get M2 ＜ 3 from △ 24-8m2 ＞ 0. Let m (x1, Y1), n (X2, Y2), ｜ X1 + x2 = - 43M ∵ f (1, 0), ｜ 1 = X1 + x2 + xm3 = - 4m9 ∵ M = - 94, not satisfy M2 ＜ 3, so the linear l equation does not exist
It is known that the center of ellipse C is at the origin of coordinates, the right focus is f (1,0), A.B is the left and right vertex of ellipse C, and P is the moving point of ellipse C different from A.B,
The maximum area of APB of triangle is 2 times root sign 3
1, find the elliptic equation (x2 / 5 + Y2 / 4 = 1)
2. The intersection of line AP and line x = 2 at point d proves that the circle with diameter BD is tangent to line PF
In fact, this question is not difficult, the key is that your first answer is wrong, the correct answer should be x ^ 2 / 4 + y ^ 2 / 3 = 1
Given that the line L: x = - 1, the point F (1,0) takes F as the focus, and l is the corresponding directrix, a vertex of the minor axis of the ellipse (whose center is not at the coordinate origin) is B,
P is the midpoint of FB, (1) find the trajectory equation of point P, and explain what curve it is, (2) take M (m, 0) as the fixed point, find the minimum value of PM, (specific steps), L
(1) Let the coordinates of point p be (x, y), then the coordinates of point B be (2x-1,2y). According to the meaning of the title, BF / ｜ 2x-1 - (- 1) ｜ = (2x-1) - 1 / ｜ BF = e, that is, (2x-2) 2 + 4y2 = 2X (2x-2), ｜ y2 = X-1 (x ＞ 1), so the trajectory of point P is a parabola (non vertex) with (1,0) as vertex, X axis as symmetry axis and opening to the right
(2) PM = √ [(x-m) ^ 2 + y ^ 2] = √ [x ^ 2 - (2m-1) x + m ^ 2-1] = √ {[x - (2m-1) / 2] ^ 2 + m-5 / 4} (x > 1). When (2m-1) / 2 > 1, i.e. m > 3 / 2, PM min = √ [(4m-5) / 2], when (2m-1) / 2 ≤ 1, i.e. m ≤ 3 / 2, PM has no minimum value
It is known that the eccentricity of the ellipse C with the center at the origin and the focus on the x-axis is 1 / 2 and passes through the point (- 1,3 / 2)
1. Find the equation of ellipse C
2. If the line L passing through the point P (2,1) is tangent to the ellipse C and the point m, the equation of the line L and the coordinates of the point m are obtained
1. First ask me to find out the equation of ellipse C: X ratio 4 + y ratio 3 = 1,
2. My method is different from that on the Internet, but I think it's right. But the answer is different from that on the Internet
(let me talk about my idea): since the line L is tangent to the ellipse C, it must be the only point m, so it is on the ellipse C. Then I set the coordinates of the point m as (x1, Y1), so the ratio of X1 to 4 + Y1 to 3 = 1. And because of the nature of the ellipse, the sum of the distances from a point on the ellipse to two focal points is 2A. So | MF1 | + | MF2 | = 2A. Because of the above problem, we find a = 4, so 2A = 4
When the two equations are simultaneous, I find X1 = 4, but when I bring in X1 ratio 4 + y ratio 3 = 1, I find Y1 = - 9. How can I get a negative number? But I calculate 5 and 6 sides, and the calculation is no problem,
1 / x ^ 2 / 4C ^ 2 + y ^ 2 / 3C ^ 2 = 1 substitute (- 1,3 / 2) into 1 / 4C ^ 2 + 3 / 4C ^ 2 = 1C ^ 2 = 1x ^ 2 / 4 + y ^ 2 / 3 = 1. You have done it right. 2 / P point is outside the ellipse, you can make two tangents, so m is not unique
RELATED INFORMATIONS
- 1. If P (2,2) is the midpoint of AB, then the equation of parabola C is______ ．
- 2. The left focus of the ellipse is (√ 3,0), and the right vertex is d (2,0). Let a (1,1 / 2) be a moving point on the ellipse. If P is a moving point on the ellipse, find the trajectory equation of the midpoint m of the line PA
- 3. Given that the distance from a point P on the parabola y2 = 16x to the x-axis is 12, then the distance from P to the focus f is equal to 12______ ．
- 4. As shown in the figure, the straight line AB; CD intersects at the point O, OE, bisecting the angle AOC, and the angle BOC - ∠ BOD = 18 ° to find the degree of ∠ BOE
- 5. As shown in the figure, points a, O and B are on the same straight line, ∠ AOC = 1 / 2 ∠ BOC + 30 °, OE bisects ∠ BOC, and calculates the degree of BOE I don't know how to draw or dictate a straight line. The right side is B and the left side is a. there is o point in the middle, with o as the point, extending from the upper right corner to e and from the upper left corner to C~
- 6. As shown in the figure, it is known that OE is the bisector of ∠ AOB, and C is a point in ∠ AOE. If ∠ BOC = 2 ∠ AOC, ∠ AOB = 114 °, calculate the degree of ∠ BOC. ∠ EOC
- 7. As shown in the figure, the straight lines AB and CD intersect at the points o, OE and of, which are bisectors of ∠ AOC and ∠ AOD respectively. It is known that ∠ BOC = 70 ° to find the degree of ∠ DOF and ∠ Coe
- 8. As shown in the figure, it is known that OD bisects BOC and OE bisects AOC, if BOC = 70 ° and AOC = 50 ° (1) Calculate the degree of AOB and its complement angle (2) Request the degree of ∠ doc and ∠ AOE, judge whether ∠ DOE and ∠ AOB are complementary, and explain the reason The DOA is a right angle, 90 degrees, and the diagram is drawn by myself
- 9. Given the angle AOB = 130 degrees, OE bisecting angle BOC, of bisecting angle AOC, calculate the degree of angle EOF
- 10. As shown in the figure: the parabola intersects the x-axis at two points a (- 1,0) and B (3,0), and intersects the y-axis at points c (0, - 3). Let the vertex of the parabola be d (1) find the analytical formula of the parabola and the coordinates of vertex D: (2) prove that the center of the circumscribed circle of △ BCD falls at the midpoint of the BD side: (3) if point P is a moving point on the coordinate axis, it can make the triangle with P, a, C as the vertex similar to △ BCD? If it exists, Please write the coordinates of point P: if not, please explain the reason
- 11. Let a straight line passing through the focus of the parabola y2 = 2px (P > 0) intersect with the parabola. Let the coordinates of the two intersections be a (x1, Y1) and B (X2, Y2) respectively. Prove: (1) y1y2 = - P2 (2) x1x2 = p24
- 12. How to find the directrix and focus of parabola? How to find the known parabola, collimator and focus? It is how to deduce the focal point and collimator of a parabola, or how to deduce a parabola from the focal point and collimator, just as the elliptic standard equation is derived by arithmetic and algebra.
- 13. Let a (x1, Y1), B (X2, Y2), then x1x2 =? Let a (x1, Y1) and B (X2, Y2) pass through the focus of the parabola (x ^ 2) = 4Y to make the chord AB, then what is X1 times x2? There is a process or a graph.
- 14. If the distance between a point P and its focus on the parabola x ^ 2 = 4Y is 2, then the distance between point P and its directrix is 2
- 15. If the distance from a point m on the parabola x ^ 2 = - 16y to the focus is 6, then the coordinate of M is
- 16. Given that the distance from the point P on the parabola x2 = 4Y to the focus is 10, then the coordinate of point P is 0______ ．
- 17. If a ＜ 0, the vertex of the parabola y = x * + 2aX + 1 + 2A * is in the? Th quadrant
- 18. When a
- 19. It is known that the vertex m of the parabola y = AX2 + BX + C with downward opening is in the second quadrant And through the points a (1,0), B (0,1) (1) Please judge the value range of real number a and explain the reason (2) Let the other intersection of the parabola and the x-axis be c. when the area of △ AMC is 25 / 16 times of the area of △ ABC, a is obtained
- 20. Generally, we can use formula to find the vertex and symmetry axis of parabola y = ax ^ 2 + BX + C (a ≠ 0). Y = ax ^ 2 + BX + C = a [x + (B / 2a)] ^ 2+ Generally, we can find the vertex and symmetry axis of parabola y = ax ^ 2 + BX + C (a ≠ 0) y=ax^2 + bx + c =a[x+(b/2a)]^2 + (4ac-b^2)/4a, y=ax^2 + bx + c It should not be converted to (x + B / 2a) ^ 2 = B ^ 2-4ac / 4A ^ 2. How can it be converted to the above form?