![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
It is known that the eccentricity of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is the root 5 / 3, and the distance from one end point of the minor axis to the right focus is 3
In an ellipse, the distance from the endpoint of the minor axis to the focus is a, so a = 3,
Because of the centrifugal rate e = C / a = √ 5 / 3, C = √ 5,
Then a ^ 2 = 9, B ^ 2 = a ^ 2-C ^ 2 = 4,
So the equation of the ellipse is x ^ 2 / 9 + y ^ 2 / 4 = 1
The ellipse focus is on the y-axis, the eccentricity e = root 3 / 2, and the shortest distance from the focus to the ellipse is 2-radical 3
The shortest distance from focus to ellipse
=Distance from focus to vertex
∴a-c=2-√3
e=√3/2
c/a=√3/2
∴a=2
c=√3
B=1
The equation of ellipse x ^ 2 / 4 + y ^ 2 = 1
Length of long axis = 2A = 4
Focal length = 2C = 2 √ 3
If you agree with my answer, please click "accept as satisfactory answer", thank you!
The eccentricity of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is the root 6 / 3, and the distance from one end point of the minor axis to the right focus is root 3, (1) solve the elliptic equation 2) Let the line L and the ellipse C intersect at two points AB, and the distance from the coordinate origin o to the straight line L is 3 / 2 of the root, and the maximum area of the triangle AOB is obtained A more detailed process is required Thank you. The key is the second question
If C / a = √ 6 / 3, a = √ 3, then C = √ 2, so the elliptic equation is: x? 2 / 3 + y? 2 / 1 = 1
It is known that the eccentricity of the ellipse x^2/a^2+y^2/b^2=1 is 2 / 3, and the distance from one end of the minor axis to the right focus is 2 (1) find the standard equation of the ellipse (2) If P is a moving point on the ellipse, F1 and F2 are the left and right focus of the ellipse respectively. Find the maximum and minimum value of vector Pf1 point multiplication vector PF2
(1) It is easy to know that the elliptic equation is x ^ 2 / 4 + y ^ 2 = 1
(2) Let Ф be the angle between vector Pf1 and PF2, then Pf1 · PF2 = | Pf1 | * | PF2 | cos Ф
According to the first definition of ellipse and cosine theorem, | Pf1 | * | PF2 | = 2B ^ 2 / (1 + cos Ф) = 2 / (1 + cos Ф)
So Pf1 · PF2 = 2cos Ф / (1 + cos Ф)
Obviously, when p is at the top of the minor axis, it is easy to know that when p is at the top of the minor axis, it is easy to know that when p is at the top of the long axis, Ф reaches the minimum, and when p is at the top of the long axis, Ф reaches the minimum, and the variation range of Ф is [0 ° and 120 °]
When Ф = 90 °, Pf1 · PF2 = 0
When Ф ≠ 90 °, Pf1 · PF2 = 2cos Ф / (1 + cos Ф) = 2 / [(1 / cos Ф) + 1], where - 1 / 2 ≤ cos Ф ≤ 1 (COS Ф ≠ 0)
So (Pf1 · PF2) max = 1, (Pf1 · PF2) min = - 2
It is known that one vertex of the ellipse is a (0, - 1), and the focus is on the x-axis. If the distance between the right focus and the straight line X-Y + 2, 2 = 0 is 3 1. Find the equation of ellipse 2. Let | m, where | m | m | m | m | is not equal to | m | m | where | K | m | is not equal to | m |
(1) If the right focus f (C, 0) is set, the distance from point F to the straight line d = (c + 2 √ 2) / √ 2 = 3
Then the equation of ellipse is x ^ 2 / 3 + y ^ 2 / 1 = 1
(2) Let m (x1, Y1) n (X2, Y2) substitute y = KX + m into x ^ 2 / 3 + y ^ 2 / 1 = 1
(3K ^ 2 + 1) x ^ 2 + 6kmx + 3M ^ 2-3 = 0 from Veda theorem X1 + x2 = - 6km (3K ^ 2 + 1)
X1 · x2 = (3m ^ 2-3) / (3K ^ 2 + 1) get Y1 Y2 from the linear equation, and then convert the condition by the distance formula between two points. It's too troublesome to fight. You can do it yourself according to this idea. After all, you can't improve only by looking at the answer in mathematics, but you have to do more and think more
Let's know that one vertex of the ellipse is a (0, - 1), the focus is on the x-axis, and the right focus is on the line X-Y + 2 If the distance between 2 = 0 is 3, try to find the elliptic equation
Let the right focus f (C, 0), (c > 0),
Then | C + 2
2|
2＝3，∴c＝
2．
∵ one vertex of the ellipse is a (0, - 1),
∴b=1，a2=3，
The elliptic equation is x2
3+y2＝1．
It is known that a vertex of an ellipse is a (0, - 1) and the focus is on the x-axis. If the distance between the right focus and the root of the line X-Y + 2 is 2 = 0, the distance is 3 Let the ellipse and the straight line y = KX + m (k is not equal to 0) intersect at different two points m and N. when | am | = | an |, the value range of M is calculated I'm in a hurry. I'll hand in my homework tomorrow
This is how the problem is done
If the right focus is set to F2 (C, 0), then | C + 2 √ 2 | / √ 2 = 3, the solution is C = √ 2,
And a (0, - 1) is the vertex, so B = 1, so a = √ 3,
Thus, the elliptic equation is: x 2 + 3 y 2 = 3 (1)
Let m (x1, Y1), n (X2, Y2), Mn midpoint Q (x0, Y0), then
x1²+3y1²=3…… II.
x2²+3y2²=3…… 3.
③ - 2, we get: (x2-x1) (x2 + x1) + 3 (y2-y1) (Y2 + Y1) = 0 4.
But x2 + X1 = 2x0, Y2 + Y1 = 2y0, (y2-y1) / (x2-x1) = k, substituting into ④
x0+3ky0=0…… 5.
Because am = an, AQ ⊥ Mn, so (Y0 + 1) / x0 = - 1 / K, i.e
x0+ky0+k=0…… 6.
⑤ (6) simultaneous solution: Q (- 3K / 2,1 / 2)
Substituting y = KX + m, we get: 1 / 2 = - 3k2 / 2 + M
m=(3k²+1)/2…… ⑦
Because q is inside the ellipse,
So (- 3K / 2) 2 + (1 / 2) 2
It is known that a vertex of an ellipse is a (0, - 1) and its focus is on the x-axis. The equation of ellipse is obtained when the right focus is from the line X-Y + 2 root sign 2 = 0 and the distance is 3
Because the focus is on the x-axis, let the right focus be (C, 0) and the distance from the line to be 3. From the formula of the distance between the point and the line, C = root 2. Because the vertex is a, so B = L, a = root 3, so the equation is x square divided by 3 + y square divided by 1 = 1
It is known that a vertex of an ellipse is a (0, - 1), and the focus is on the x-axis. If the distance from the right focus to the line X-Y + 2 √ 2 = 0 is 3 (√ is the root sign) Given that a vertex of an ellipse is a (0, - 1), and the focus is on the x-axis, if the distance from the right focus to the straight line X-Y + 2 √ 2 = 0 is 3 (√ is the root sign) (1) find the focus coordinates of the elliptic equation (2) let the line L: y = x + m, whether there is a real number m
1. Let the right focus coordinate be F2 (C, 0),
Straight line: X-Y + 2 √ 2 = 0,
According to the point line distance formula, the distance from F2 to the straight line D: 1
d=|c-0+2√2|/√(1+1)=|c+2√2|/√2=3,
∵c>0.
∴c+2√2=3√2,
c=√2,
A (0, - 1) is the lower vertex of the minor axis,
∴b=1,
a=√(b^2+c^2)=√3,
The elliptic equation is: x ^ 2 / 3 + y ^ 2 = 1
The second question is incomplete
It is known that OA ⊥ OC, and ∠ AOB: ∠ AOC = 2:3, then the degree of ∠ BOC is () A. 30° B. 150° C. 30 ° or 150 ' D. 90°
∵OA⊥OC，
∴∠AOC=90°，
∵∠AOB：∠AOC=2：3，
∴∠AOB=60°．
Because there are two kinds of position of ∠ AOB: one is inside ∠ AOC, the other is outside ∠ AOC
① When in ∠ AOC, ∠ BOC = 90 ° - 60 ° = 30 °;
② When it is outside of AOC, BOC = 90 ° + 60 ° = 150 °
Therefore, C
RELATED INFORMATIONS
- 1. Given that OA is perpendicular to OC, B is in the plane, angle AOB: angle AOC = 2:3, what is the degree of angle BOC Can you explain the white point
- 2. There are three rays OA, OB and OC in the same plane. For example, the angle BOC is 5 degrees smaller than 2 / 5 of the complementary angle of angle AOB, and the angle AOC is 10 degrees smaller than the complementary angle of angle BOC There must be a line segment relation There are three cases to be discussed by classification
- 3. Two circles with the same radius intersect, and the distance between the centers of the two circles is exactly equal to the radius of 10 cm
- 4. The distance from each point on the circle to the center of the circle is equal to______ The points whose distance from the center of the circle is equal to the radius are all in the______ .
- 5. If the distance between the centers of the two circles is d = 8, and the lengths of the radii of the two circles are the two roots of the equation x ^ - 7x + 12 = 0, then what is the positional relationship between the two circles? Please explain the reason! In a hurry
- 6. It is known that the radius of ⊙ o is 5cm, the point P is the outer point of ⊙ o, Op = 8cm. A circle with P as the center is tangent to ⊙ o, and the radius of the circle is______ cm．
- 7. Find the chord length cut by the square of circle x + the square of Y-2 times the root sign 2x-4y-2 = 0 on the x-axis
- 8. Given that the radius of a circle is 5 and the length of a chord is 8, what is the distance between the midpoint of the chord and the midpoint of the arc corresponding to the chord
- 9. Given that the radius of circle O is ob = 5cm, chord AB = 6cm, D is the midpoint of arc AB, the length of chord BD is calculated
- 10. As shown in the figure: in a circle with a diameter of 10 cm, the distance from the center of the circle to the chord AB is 4cm
- 11. If there is no real root in the quadratic equation x ^ 2 + (M + 2) x + (1 / 4) m ^ 2 = 0, the root number m ^ 2 + │ M-1 │
- 12. Over the point（ 3, − 2) passes through the center of circle x2 + y2-2y = 0, then the inclination angle of line L is () A. 30° B. 60° C. 150° D. 120°
- 13. If the equation of the circle is the square of X + the square of Y - 2x + 4y-20 = 0, find the linear equation of the diameter of the origin
- 14. Find the distance between the parallel lines 3x-2y-5 = 0 and 6x-4y + 3 = 0
- 15. It is known that the radii R1 and R2 of the two circles are two of the equations x2-7x + 10 = 0, and the distance between the centers of the two circles is 7, then the position relationship between the two circles is______ .
- 16. As shown in the figure, in the plane rectangular coordinate system xoy, the vertex e coordinates of rectangular oefg are (4,0), and the vertex g coordinates are (0,2). Rotate the rectangular oefg anticlockwise around point O, so that point F falls on the point n of the Y axis, and the rectangle omnp, OM and GF intersect at point a (1) Judge whether △ OGA and △ NPO are similar and explain the reasons; (2) The analytic formula of inverse proportional function of point a is obtained; (3) If the image of the inverse scaling function obtained in (2) intersects with EF at point B, please explore whether the line AB is perpendicular to OM and explain the reason
- 17. Given the point a (- 4,0) B (2,0), if the point is on the image of the linear function y = - 2 / 1X + 2 and the triangle ABC is a right triangle, several conditions of point C are satisfied
- 18. In the plane rectangular coordinate system x0y, the coordinates of the parallelogram A and B are (2,3), (5,1), respectively (1) If point C is on the positive half axis of the X axis and D is on the positive half axis of the Y axis, write the coordinates of the vertices C and D (2) If (1) condition is changed to point C and D on the coordinate axis, write the coordinates of C and D Ask God to answer, correctly add points
- 19. In the rectangular coordinate system xoy, point B and point a (- 1,1) are symmetric about the origin o, P is a moving point, and the product of slope of straight line AP and BP is equal to - 1 / 3 Drawing and finding that APB and MPN are complementary sinAPB=sinMPN PA/PM=PN/PB.a (x0+1)/(3-x0)=(3-x0)/(x0-1).b Why can a write B
- 20. It is known that in the plane rectangular coordinate system, point P is in the third quadrant, the circle P is tangent to the x-axis, is tangent to the point Q, and intersects the y-axis with m (0, - 3) n（