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Given the coordinates of the focus of the ellipse and a point P that the curve passes through, how can we find the standard equation of the ellipse? It is known that the two focal coordinates of an ellipse are F1 (- √ 3,0) F2 (√ 3,0) and pass through the ellipse standard equation of point P (√ 5, √ 6)
Given the coordinates of the focus of the ellipse and a point P that the curve passes through, how can we find the standard equation of the ellipse?
A: use the undetermined coefficient method
That is, from the known ellipse focus coordinates,
Let the elliptic standard equation satisfy the condition
And then by the condition that the curve passes through a point P,
Then the coordinates of the point P should satisfy the elliptic standard equation,
The coordinates of the point P are substituted into the elliptic standard equation,
Then we can get an equation about the coefficients a and B to be determined, which is called equation (1);
From the known ellipse focus coordinates,
Another equation about the coefficients a and B to be determined is obtained, which is called equation (2);
Simultaneous equations (1) and (2),
Solve the equations to obtain the value of the square of the undetermined coefficients a and B (that is),
We substitute the square of a and B into the standard elliptic equation,
It is the standard equation of the ellipse
For example:
It is known that the two focal coordinates of an ellipse are F1 (- √ 3,0) F2 (√ 3,0) and pass through the ellipse standard equation of point P (√ 5, √ 6)
The two focal coordinates of the ellipse are F1 (- √ 3,0) F2 (√ 3,0)
We know that the focus of the ellipse is on the x-axis,
Therefore, the standard equation of ellipse should be set as follows:
x²/a²+y²/b²=1
From the condition that the ellipse passes through the point P (√ 5, √ 6), we get
5/a²+6/b²=1 (1)
Then the two focal coordinates of the known ellipse are F1 (- √ 3,0) F2 (√ 3,0),
And a, B, C
c²=a²-b²
We get a & # 178; - B & # 178; = (√ 3) &# 178; = 3 (2)
Simultaneous equations (1) and (2),
Solve the equations to obtain the undetermined coefficient a, B, the square of the value is:
B & # 178; = 4 + radical 34 = 4 + √ 34, B & # 178; = 4 - radical 34 = 4 - √ 34
a^2-b^2=(√3)^2=3
5/a^2+6/b^2=1
=>5/a^2+6/(a^2-3)=1
=>A ^ 2 = 7 + root 34 B ^ 2 = 4 + root 34
It is known that the two focal coordinates of the ellipse are (0, - 2), (0,2) respectively, and through (3 / 2,5 / 2), the standard equation of the ellipse is obtained
The major axis is on the Y axis, C = 2
According to the definition of ellipse: √ [(3 / 2 - 0) ^ 2 + (5 / 2 + 2) ^ 2] + √ [(3 / 2 - 0) ^ 2 + (5 / 2 - 2) ^ 2] = 2A
That is: 2A = 2 √ 10
a=√10
Then: B ^ 2 = a ^ 2 - C ^ 2 = 6
The standard equation of ^ 2 / y = 1 / 6
From the focus coordinate, we can see that C = 2, and the major axis is on the Y axis
Then the distance from the passing point to the two focal points is 3 √ 10 / 2 and √ 10 / 2 respectively, so a = √ 10, so B = √ 6
So the standard equation is
x^2/6 + y^2/10 = 1
Given the ellipse focus coordinates and a point P that the curve passes through, how to find the standard equation of ellipse? Find the routine~
For example, given that the two focal coordinates of an ellipse are F1 (- 2,0) F2 (2,0) and pass through the point P (5 / 2, - 3 / 2), then the standard equation of the ellipse is
How to answer this question
In general, we first find the length of the semimajor axis A, the length of the semiminor axis B and the half focal length C of the ellipse, and then use the standard equation of the ellipse X & sup2 / / A & sup2; + Y & sup2 / / B & sup2; = 1 to write it. For example, if we know that the two focal coordinates of the ellipse are F1 (- 2,0) F2 (2,0) and pass through the point P (5 / 2, - 3 / 2), then the standard equation of the ellipse is F1 (- 2,0) F2 (2,0)______ .|PF...
Given the focal coordinates and a, the standard elliptic equation can be solved
The focal coordinates are (0, - 4) (0,4), a = 5. Find the elliptic standard equation suitable for the following conditions
It's very simple. A = 5, C = 4, so B ^ 2 = a ^ 2-C ^ 2 = 9, so B = 3
The focus is on the X axis, so the elliptic equation ((x ^ 2) / 25) + ((y ^ 2) / 9) = 1
⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙___ .
From the two intersection points of the straight line and hyperbola about the Y axis symmetry, we can get k = 0, that is, the linear equation is y = 1; the hyperbolic equation is x2-y2-y-9 = 0. Two simultaneous analytical expressions are y = 1x2-y2-y-9 = 0, the solution is x = 11y = 1 or X = - 11y = 1, so the intersection coordinates are (11,1) or (- 11,1)
Sin (Wu + a) = - why is Sina negative?
Suppose a is the first quadrant angle, then Wu + A is the third quadrant angle
Sin is positive in the first quadrant and negative in the third quadrant
It's understandable
In the plane rectangular coordinate system, the vertex o of the rectangular oacb is at the origin of the coordinate, the vertices a and B are respectively on the positive half axis of the X axis and Y axis, OA = 3, OB = 4, and D is the midpoint of the edge ob. (1) if e is a moving point on the edge OA, when the perimeter of △ CDE is the smallest, calculate the coordinates of point E; warm tips: as shown in the figure, you can make a symmetrical point d 'of point d about the upper axis, connect CD' and X axis to point E, and then (2) if e and F are two moving points on the edge OA, and EF = 2, when the perimeter of the quadrilateral cdef is minimum, the coordinates of points E and F can be obtained
(1) Let d be the symmetric point d 'of point d about X axis and connect CD' with X axis at the point E. ∵ ob = 4, OA = 3, D is the midpoint of ob, ∵ od = 2, then the coordinates of D are (0,2), C are (3,4) ∵ the coordinates of d 'are (0, - 2)
The given line y = 2x + 1. (1) find the coordinates of the intersection a of the given line and Y axis; (2) if the line y = KX + B and the known line are symmetric about y axis, find the values of K and B
(1) When x = 0, y = 1, so the coordinates of the intersection a of the line y = 2x + 1 and the Y axis are (0, 1); (2) for the line y = 2x + 1, when x = 0, y = 1; when y = 0, x = - 12, that is, the intersection points of the line y = 2x + 1 and the two coordinate axes are (0, 1), (- 12, 0), ∵ the two lines are symmetrical about the Y axis ∪ the straight line y = KX + B passes through the point (0, 1), (12, 0), so 1 = B0 = 12K + B, ∪ k = − 2B = 1 So k = - 2, B = 1
Sin (3 + a) = - Sina
Why?
sin(3π+A)
=sin(2π+π+A)
=sin(π+A)
=-sinA.
sin(2π+π+A)=sin(π+A)=-sinA
In the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length 2 are respectively on the positive half axis of Y axis and X axis, and the point O is at the origin. Now rotate the square oabc clockwise around the point O, and stop rotating when the point a falls on the line y = x for the first time. In the process of rotation, AB intersects the line y = x at the point m, BC intersects the X axis at the point n (as shown in the figure). (1) calculate the direction swept by the side OA in the process of rotation (2) in the process of rotation, when Mn and AC are parallel, calculate the degree of rotation of the square oabc; (3) if the perimeter of △ MBN is p, does the value of P change in the process of rotating the square oabc? Please prove your conclusion
(1) The angle between the line y = x and the Y axis is 45 ° and the OA rotates 45 ° for the first time. The area swept by OA during the rotation is 45 π × 22360 = π 2. (2) ∵ Mn ∥ AC, ∵ BMN = ∵ BAC = 45 ° and ∵ BNM = ∵ BCA = 45 ° In the rotation process, when Mn and AC are parallel, the rotation degree of oabc is 45-22.5 ° = 22.5 °. (3) there is no change of P value in the rotation process of oabc. It is proved that when the Y-axis of Ba intersection is extended to e point, ∠ AOE = 45 ° - AOM, ∠ con = 90 ° - 45 ° - AOM = 45 ° - AOM, ∠ AOE = con C. In the process of rotating the square oabc, the value of P does not change
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