In the plane rectangular coordinate system, the coordinates of a point a are (a, b). (1) the distance from a to the origin o is expressed by the quadratic radical; (2) the point B (root 5, - radical 3) is calculated Distance to origin

In the plane rectangular coordinate system, the coordinates of a point a are (a, b). (1) the distance from a to the origin o is expressed by the quadratic radical; (2) the point B (root 5, - radical 3) is calculated Distance to origin

Root sign (a 2 + B 2)
Root number (5 2 + | - 3|) = root 34

It is known that an ellipse in the plane rectangular coordinate system xoy has its center at the origin, the left focus is f (- radical 3,0), the right vertex is d (2,0), and the point a (1 It is known that an ellipse in the plane rectangular coordinate system xoy has its center at the origin, the left focus is f (- radical 3,0), and the right vertex is d (2,0). Set point a (1,1 / 2). (1) find the standard equation of the ellipse. (2) if P is a moving point on the ellipse, find the trajectory of the end point m of the line PA

1、a=2,c=√3,b=1,
∴x^2/4+y^2=1.
2. Should it be the midpoint?
Parameter equation: x = 2cost, y = Sint,
P(2cost,sint),
x=(1+2cost)/2,
cost=(2x-1)/2,(1)
y=(1/2+sint)/2,
sint=(2y-1/2),(2)
(x-1/2)^2+4(y-1/4)^2=1.

It is known that the midpoint of the ellipse is at the origin, the left focus is f (- radical 3,0), the right vertex is d (2,0), and the point a (1,1 / 2) is set (1) The standard equation of ellipse (2) If P is a moving point on an ellipse, find the trajectory equation of midpoint m of line PA (3) A straight line L with slope 1 passes through the right focus of the ellipse and intersects the ellipse at points B and C to find the length of chord BC

(1) So the standard equation of ellipse is x ^ 2 / 4 + y ^ 2 = 1 (2) if the coordinates of midpoint m are (x, y), then the coordinates of point P are 2x-1,2y-1) and the point P is on the ellipse, so the equation x ^ 2 + 4Y ^ 2-x-2y = 0 (3) BC = 2a-eix1 + x2ia = 2 e = √ 3 / 2 straight line L = y = (x -...)

It is known that an ellipse in the plane rectangular coordinate system xoy has its center at the origin, the focal point is (- radical 3,0), and the point a (1,2) is set at the right vertex D (2,0) (1) Find the standard equation of the ellipse (2) if P is a moving point on the ellipse, find the trajectory equation of the midpoint m of the line PA (3) the intersection ellipse of the straight line passing through the origin o at points B, C, and find the maximum area of the triangle ABC only needs to answer question 3

Well, just do what you want
1) The elliptic equation is obtained as follows
3) Let the straight line be y = KX 2
① (2) the relationship of X1 + x2 and x1x2 is obtained
According to the distance formula between point a and line BC, the height h on BC is obtained
The length of BC can be obtained from the above
Then s = H / 2 * BC
There will be K in the limit, we can discuss it. If we list the formula, we will not find the range of area
Contact me then
That's the way of thinking. Everything changes

The known point (0, - radical 5) is a vertex of an ellipse whose center is at the origin and the major axis is on the x-axis. The eccentricity is 6 / 6, and the left and right focus of the ellipse are F1 and F2 respectively

∵ the long axis of the ellipse is on the x-axis, and the point (0, - 5) is a vertex of the ellipse ᙽ B = √ 5 ① And ∵ the centrifugal rate is √ 6 / 6 ᙽ e = C / a = √ 6 / 6 ② And ∵ a ∵ B ∵ C ③ The standard equation of ellipse is: x? 2 / 6

The right focus of hyperbola C with center at origin is (2,0) and right vertex is (radical 3,0) If the straight line y = KX + m (K ≠ 0, m ≠ 0) and hyperbola intersect at different two points m and N, and the vertical bisector of segment Mn crosses the point (0, - 1), the value range of real number m is calculated

∵c=2,a=√3
The hyperbolic equation is x? 2 / 3-y? 2 = 1
set up
If the slope of CD = k, then the slope of the vertical bisector = - 1 / K,
Let C and d be (x1, Y1), (X2, Y2), and let m in CD be (a, b),
Let B 2 be a bisector
Because l passes through (0, - 1)
B2 = - 1
L is y = - X / k-1
Because (x12-x22) / 3 = (Y12 + 1) - (Y22 + 1)
=>(x1+x2)/3(y1+y2)=(y1-y2)/(x1-x2)=k
Then a / 3B = K,
If m point is also on the line L, then B = - A / k-1 (replace k = A / 3b)
B = - 1 / 4, k = - 4A / 3
Obviously, point m is also on the line y = KX + m, then B = Ka + M
Then - 1 / 4 = - 3k2 / 4 + M
3k2=4m+1
Substituting y = KX + m into hyperbolic equation to eliminate y
X2 / 3-k2x2-2kmx-m2-1 = 0 if the equation has two real roots
Then 4m2k2-4 (- M2-1) (1 / 3-k2) > 0
=>m2/3-k2+1/3>0
=>m2+1>3k2=4m+1

In the plane rectangular coordinate system, the focal length of the ellipse x2 / A2 + Y2 / b (a > 0, b >) is 2 In the plane rectangular coordinate system, if the focal length of the ellipse x2 / A2 + Y2 / b (a > 0, b >) is 2, the center of the circle O, a is the radius of the circle, and the two lines passing through the point (A2 / C, 0) are perpendicular to each other, then the eccentricity E=

2c=2
C=1
(A2 / C, 0) can be recorded as a = (a2,0)
The angle between tangent L and X axis is 45 °,
Angle Bao = 45 °,
OB=a,
The triangle Bao is an isosceles right triangle,
AO=√2*OB=√2a.
The abscissa of point a is A2
a2=√2a,a=√2.
e=c/a=√2/2.

In the plane rectangular coordinate system xoy, it is known that the eccentricity of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is √ 2 / 2, and its focus is on the circle x ^ 2 + y ^ 2 = 1, (1) find the ellipse

1. The focus of the ellipse is on the x-axis, and the eccentricity e = C / a = √ 3 / 2
Let C = √ 3k, a = 2K, K

In the plane rectangular coordinate system, the ellipse x2 a2+y2 The focal length of B2 = 1 (a > b > 0) is 2c, with o as the center of the circle and a as the radius of the circle c. If the two tangent lines of the circle are perpendicular to each other, the eccentricity e = () A. Two Two B. 2 C. Three D. Two

Method 1: as shown in the figure, tangent PA and Pb are perpendicular to each other,
The radius OA is perpendicular to pa,
So △ OAP is an isosceles right triangle,
A2
C=
2a．
The solution is e = C
A=
Two
2．
Then the centrifugal rate E=
Two
2；
Method 2: the eccentricity of the key ellipse is less than 1,
Analysis option, only a is less than 1,
Therefore, a

The ellipse C with the origin o of plane rectangular coordinate system as the center passes through point a (2,3) and the right focus is f (2,0)

The elliptic standard equation is x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
a^2-b^2=c^2=4
By substituting point a (2,3) into the elliptic equation, the
4b^2+9a^2=a^2b^2
4b^2+9(4+b^2)=(4+b^2)b^2
The result shows that B ^ 2 = 12, a ^ 2 = 16
Therefore, the elliptic equation is: x ^ 2 / 16 + y ^ 2 / 12 = 1