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In the plane rectangular coordinate system, there is an ellipse with F1 (0, - radical 3) and F2 (0, radical 3) as the focus, and the eccentricity is half of the root 3
The ellipse of root 3 is obviously a = 2, C = √ 3, B = 1, and the elliptic equation is x / 4 + Y / 1 = 1; in the part of the first quadrant of the ellipse, set P point as (x0, Y0) y '= - x0 (2 √ (4-x0)) as the tangent slope of the tangent point, y-y0 = - x0 (2 √ (4-x0)) as the tangent slope of the tangent point, y-y0 = - x0 (2 √ (4-x0)) * (x-x0) is the tangent equation, so, point a is (4 / x0, 0, B is (4 / x0,0), similarly, point B is (0 0 0 0), similarly, B is (0 0 0 0 0 0), similarly, B is (0) for point B is (, 1 / Y0), Let x = 4 / x0, 1 / x = x0 / 4, the same reason 1 / y = Y0, because the ellipse satisfies x0 / 4 + Y0 / 1 = 1; (x0 / 4 * 2) + Y0 = 1; (x0 / 4 * 2) + Y0 = 1; -- > (2 / x) + (1 / y) = 1 is the trajectory equation m (4 / x0,1 / Y0); x0 / 4 + Y0 / 1 = 1; another x0 = 2cosa; Y0 = Sina; om = (2 / COSA) + (1 / Sina) = Sina; om = (2 / COSA) + (1 / Sina) = 1; (4 / 4 + Y0 / 1 / y = 1; another x0 = 2cosa; Y0 = Sina; om = (2 / COSA) + (1 / Sina) = 1 / Sina 2-T / T (1-T) (t = COSA) = u (0 < T < 1) the maximum value u > = 1 / 2 can be obtained by using the discriminant formula, | om | > = √ 2 / 2
It is known that the two foci of the ellipse are F1 at (- radical 3,0), F2 (radical 3,0) eccentricity e = radical 3 / 2 Let the straight line L: y = x + m, if l and the ellipse intersect at two points P and Q, and | PQ | is equal to the length of the minor axis of the ellipse, calculate the value of M
C = √ 3E = C / A, so a = 2, so B = 1 x 2 / 4 + y 2 = 1y = x + m, substituting x 2 + 4Y 2 = 45x? + 8mx + (4m? 2 - 4) = 0 (x1-x2) 2 = (x1 + x2) x2 - 4x1x2 = 64M? 2 / 25 - (16m? - 16) / 5 = (- 16m? + 80) / 25y = x + my1-y2 = X1 + m-x2-m
Elliptic C passing through point P (radical 2 / 2,1) with F 1 (0, - 1), F 2 (0,1) as the focus 1. Find the equation of elliptic C 2. The moving line L passing through point s (- 1 / 3,0) intersects ellipse C at two points A. B. is there a fixed point t on the coordinate plane so that no matter how l rotates, the circle with ab as diameter will always pass through the point t? If so, find out the coordinates of point t; if not, please explain the reasons
If F 1 (0, - 1) and F 2 (0,1) pass through point P (radical 2 / 2,1), we know that the focus of ellipse is on Y axis. We can set the standard equation as follows: x-square / b-square + y-square / A-Square = 1 A-Square = b-square + c-square
It is known that the focal points of ellipse C are F1 (- 2) 2, 0) and F2 (2 Let the line y = x + 2 intersect ellipse C at two points a and B. find the coordinates of the midpoint of line ab
Let the equation of elliptic C be x2
a2+y2
b2=1，
A = 3, C = 2
2，
B=
A2 − C2 = 1. (3 points)
The equation of ellipse C is x2
9 + y2 = 1. (5 points)
Simultaneous equations
y＝x+2
X2
9 + y2 = 1, eliminating y leads to 10x2 + 36x + 27 = 0,
Because the discriminant of the quadratic equation △ > 0, there are two different intersections between the straight line and the ellipse, (9 points)
Let a (x1, Y1), B (X2, Y2), then X1 + x2 = - 18
5，
Therefore, the midpoint coordinate of line AB is (- 9)
5，1
5) (12 points)
It is known that the center of hyperbola is at the origin, the focal points F1 and F2 are on the coordinate axis, the eccentricity is root 2, and the point m (3, m) passing through points (4, - 10) is on the hyperbola (1) Solving hyperbolic equation (2) Prove that vector MF1 times vector MF2 = 0 (3) Calculate the area of △ f1mf2
(1) Let the focus be on the X axis, and the hyperbolic equation is: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1,
c/a=√2,(a^2+b^2)=2a^2,a=b,
x^2/a^2-y^2/a^2=1,
The hyperbola passes through the point (4, - 10),
Put it into the equation, a = √ 6,
The hyperbolic equation is: x ^ 2 / 6-y ^ 2 / 6 = 1, which is the real axis on the X axis,
If the real axis is on the y-axis, there is no real number solution for point (4, - 10), so the focus cannot be on the y-axis
(2) And m (3, m) on hyperbola,
M = ± √ 3, C = EA = √ 2 * √ 6 = 2 √ 3, focal coordinates: F1 (- 2 √ 3,0), F2 (2 √ 3,0),
Vector f1m = {3 + 2 √ 3,3}, vector F2m = {3-2 √ 3,3},
Vector f1m · vector F2m = (3 + 2 √ 3) I · (3-2 √ 3) I + 3j · (3-2 √ 3) I + (3 + 2 √ 3) I · 3j + 3j · 3j = - 3 + 3 = 0,
Here I and j are the unit components of horizontal and vertical vectors, and the dot product of I and j is 0,
Similarly, when m = = - √ 3, the result is the same,
The vector f1m · vector F2m = 0, and the two vectors are perpendicular to each other
The Pythagorean theorem can be used to prove that f1m ^ 2 + F2m ^ 2 = F1F2 ^ 2, or the reciprocal relationship between the slopes of f1m and F2m of the straight line is obtained to prove that two vectors are perpendicular, and the point product of two vectors is 0
(3) In △ mf1f2, | F1F2 | 2C = 4 √ 3, high = √ 3,
∴S△MF1F2=|F1F2|*h/2=4√3*√3/2=6.
It is known that the center of hyperbola is at the origin, the focus F1F2 is on the coordinate axis, and the eccentricity e is equal to the root 2 and passes through the point (4, - root 10) 1 find the hyperbolic equation, 2 if the point m (3, m) is on a hyperbola, prove the vector MF1 * vector MF2 = 0, 3 find s triangle f1mf2
1) Let the equation be x 2 / a 2 - y 2 / B 2 = 1
∵c²/a²=e²=2 b²=c²-a² ∴b²=2a²-a²=a²
16 / a? 10 / a? 2 = 1 = > A? 2 = 6 [if a? Is calculated as a negative number, the focus is on the y-axis]
The equation x? 2 / 6-y? 2 / 6 = 1 is obtained
2) When XM = 3, YM = M = ± √ (9-6) = ±√ 3 (i.e., YM '= √ 3; YM' '= - √ 3)
∵F1(-√12,0) ； F2(√12,0)
The slope of m'f1 K (m'f1) = (YM '- yf1) / (XM' - xf1) = (√ 3-0) / (3 + √ 12) = 2 - √ 3
The slope of m'f2 is K (m'f2) = (YM '- yf2) / (XM' - xf2) = (√ 3-0) / (3 - √ 12) = - 2 - √ 3
2 - √ 3 = - 1 / (- 2 - √ 3)
∴M'F1⊥M'F2
Similarly, M "F1 ⊥ m" F2
∴MF1⊥MF2
The dot product of vector MF1 and vector MF2 is zero
3）|F1F2|=2√12 |ym|=√3
∴S⊿F1MF2=(|F1F2|*|ym|)/2=2√12*√3/2=6
It is known that the center of hyperbola is at the origin, the focal points F1 and F2 are on the coordinate axis, and the eccentricity is root 2 and passes through the point (4, - root 10) (1) Solving hyperbolic equation (2) If the point m (3, m) is on hyperbola, it is proved that MF1 ⊥ MF2 (3) Find the area of triangle f1mf2
1) Let the equation be x 2 / a 2 - y 2 / B 2 = 1
∵c²/a²=e²=2 b²=c²-a² ∴b²=2a²-a²=a²
16 / a? 10 / a? 2 = 1 = > A? 2 = 6 [if a? Is calculated as a negative number, the focus is on the y-axis]
The equation x? 2 / 6-y? 2 / 6 = 1 is obtained
2) When XM = 3, YM = M = ± √ (9-6) = ±√ 3 (i.e., YM '= √ 3; YM' '= - √ 3)
∵F1(-√12,0) ； F2(√12,0)
The slope of m'f1 K (m'f1) = (YM '- yf1) / (XM' - xf1) = (√ 3-0) / (3 + √ 12) = 2 - √ 3
The slope of m'f2 is K (m'f2) = (YM '- yf2) / (XM' - xf2) = (√ 3-0) / (3 - √ 12) = - 2 - √ 3
2 - √ 3 = - 1 / (- 2 - √ 3)
∴M'F1⊥M'F2
Similarly, M "F1 ⊥ m" F2
∴MF1⊥MF2
The dot product of vector MF1 and vector MF2 is zero
3）|F1F2|=2√12 |ym|=√3
∴S⊿F1MF2=(|F1F2|*|ym|)/2=2√12*√3/2=6
Let F 1 be the center of the hyperbolic equation It is known that the center of hyperbola is at the origin, the focal points F1 and F2 are on the coordinate axis, the eccentricity e = (root 6 /) 2 and pass through the point (4-radical 6) (1) Find the hyperbolic equation (2) Write the Quasilinear equation and asymptote equation of the hyperbola
(1) 2 / 2 = C / a = = > 6 / 4 = C ^ 2 / A ^ 2 = = > 6 / 4 = C ^ 2 / A ^ 2 = = = > A ^ 2 = (2C ^ 2) / 3 B ^ 2 / 3 B ^ 2 = C ^ 2-A ^ 2 = (C ^ 2) / 3 set the focus on the X table. X ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 put the point (4 6) into the X table, x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 put the point (4 6) into the X table x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 put the point (4 6) on the Y table to get C ^ 2 = - 23...It's a good idea
It is known that the center of hyperbola is at the origin, the focal point f1.f2 is on the coordinate axis, the eccentricity is 2 under the root sign, and the point (4, 10 under the negative root sign) (1) is used to find the double It is known that the center of the hyperbola is at the origin, the focus f1.f2 is on the coordinate axis, and the eccentricity is 2 under the root, and it passes through the point (4, 10 under the negative root) (1) Find the hyperbolic equation
Let e = √ 2, a = B, let hyperbolic equation, and then replace the point (4, 10 under the negative root) into the equation, and get a = b = 6. Note that there are two equations. The focus f1.f2 may be x-axis or y-axis on the coordinate axis
It is known that the center of the ellipse is at the origin, the focus is F1 (0, - 2 times the root 2). F2 (0,2 times the root 2), and the eccentricity e = 2 / 3 times the root 2
∵ focus is F1 (0, + - 2 √ 2)
∴c=2√2
And ∵ e = 2 √ 2 / 3
∴a=3
The elliptic equation is x? 2 / 9-y? 2 / 8 = 0
RELATED INFORMATIONS
- 1. It is known that the center of the ellipse is at the origin, a focal point F1 (0, negative 2 times the root 2), and the eccentricity e satisfies: 2 / 3, e, 4 / 3 into an equal ratio sequence The equation of solving ellipse
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- 18. In the plane rectangular coordinate system, O is the origin of the coordinate. Two points a (3, 1) and B (- 1, 3) are known. If point C satisfies the OC=α OA+β OB, where α, β∈ R and α + β = 1, find the locus of point C and its trajectory equation
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