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If the length of an arc is 12 π cm and the center angle of the arc is 108 °, the radius of the arc is
12 π = 108 / 180 * r, then r = 20
In a circle with a radius of 15 cm, what is the central angle of the arc with a length of 5 π cm
60°
∵ C circle = 30 π cm
∴5π÷30π=1/6
360°×1/6=60°
Trajectory equation of the midpoint of each chord passing through the left focus of ellipse x ^ 2 + 4Y ^ 2 = 4
Great Xia, help to solve it!
Suppose that the coordinates of the two intersections of the chord and the ellipse are (x1, Y1) and (X2, Y2) respectively, and substituting them into the elliptic equation, it is obtained that
X1 ^ 2 + 4Y1 ^ 2 = 4 and X2 ^ 2 + 4y2 ^ 2 = 4
If you subtract the two expressions, you get
(x1+x2)(x1-x2)+4(y1+y2)(y1-y2)=0
Divide both sides by 2 (x1-x2) to get
(x1+x2)/2+4[(y1+y2)/2][(y1-y2)/(x1-x2)]=0
Where (x1 + x2) / 2 and (Y1 + Y2) / 2 are the abscissa and ordinate of the midpoint of each chord, which can be expressed by X and Y respectively
Where (y1-y2) / (x1-x2) is the slope of the chord, which can be expressed by K
Then the original formula can be written as x + 4ky = 0 and K = - X / (4Y)
The chord passes through the left focus (- root 3,0), so the slope k can also be expressed as
K = (y-0) / (x + root 3)
Two K are equal, so - X / (4Y) = (y-0) / (x + radical 3)
The trajectory equation is obtained
X ^ 2 + root 3 * x + 4Y ^ 2 = 0
The trajectory equation of the middle point of the chord of a point P (1, - 1) in the ellipse x ^ 2 + 4Y ^ 2 = 16
Let the intersection ellipse a (x1, Y1) B (X2, Y2) a B (x0, Y0) let x0 not be 1 and let the intersection ellipse y = K (x-1) - 1, then Y0 = K (x0-1) - 1. K = (Y0 + 1) / (x0-1) substitute AB into the ellipse X1 ^ 2 + 4Y1 ^ 2 = 16x2 ^ 2 + 4y2 ^ 2 = 16 to get (x1 ^ 2-x2 ^ 2) + 4 (Y1 ^ 2-y2 ^ 2) = 0 (x1 + x2) (x1-x2) + 4 (Y1
How to solve the trajectory equation of the midpoint m of the chord PQ at a point a (1,1) in the ellipse x ^ 2 + 4Y ^ 2 = 16?
Let P (x1, Y1), q (X2, Y2), then X1 & # 178; + 4Y1 & # 178; = 16 x2 & # 178; + 4y2 & # 178; = 16 get X1 & # 178; - x2 & # 178; = - 4 (Y1 & # 178; - Y2 & # 178;), that is, (x1 + x2) (x1-x2) = - 4 (Y1 + Y2) (y1-y2), let PQ midpoint coordinate be (m, n), so (y1-y2) / (x1-x2) = - (x1 + x2)
The chord that passes through the ellipse 3x square + 4Y square = 12. Find the trajectory equation of the midpoint of the chord
... square sign can't play, it's really no points, please forgive me!
Sorry for missing a left focus. It should be a chord passing through the left focus of the ellipse!
3x^2+4y^2=12
x^2/4+y^2/3=1
Left focus (- 1,0)
Let the equation of chord be y = K (x + 1) intersection a (x1, Y1) B (X2, Y2) midpoint coordinates (x, y) x = (x1 + x2) / 2, y = (Y1 + Y2) / 2
x1^2/4+y1^2/3=1
x2^2/4+y2^2/3=1
The difference is k = (y1-y2) / (x1-x2) = - 3 (x1 + x2) / 4 (Y1 + Y2) = - 3x / 4Y
Substituting k = - 3x / 4Y into y = K (x + 1)
y=(-3x/4y)(x+1)
It is reduced to 3 (a + 1 / 2) ^ 2 + 4Y ^ 2 = 3 / 4
In general, it will tell you the slope of the string!
In the ellipse x ^ 2 / 16 + y ^ 2 / 4 = 1, find the total pass of the linear equation of the chord which passes through the point m (1,1) and is bisected by this point
In the ellipse x ^ 2 / 16 + y ^ 2 / 4 = 1, the whole process of simplifying the linear equation of the chord passing through point m (1,1) and bisected by this point is obtained,
Let a (x1, Y1), B (X2, Y2)
x1^2/16+y1^2/4=1
x2^2/16+y2^2/4=1
Subtracting: (x1 + x2) (x1-x2) / 16 + (Y1 + Y2) (y1-y2) / 4 = 0
And X1 + x2 = 2 * 1 = 2, Y1 + y2 = 2
We get: (x1-x2) / 8 + (y1-y2) / 2 = 0
So the slope k = (y1-y2) / (x1-x2) = - 1 / 4
So the straight line is Y-1 = - 1 / 4 (x-1), that is y = - X / 4 + 5 / 4
In the ellipse x216 + y24 = 1, the slope of the linear equation of the chord passing through point m (2,1) and bisected by this point is ()
A. 12B. -12C. 18D. -18
Let a (x1, Y1) and B (X2, Y2) intersect the ellipse at points a and B. from the meaning of the title, we can get x1216 + y124 = 1x2216 + y124 = 1. By subtracting the two formulas, we can get (x1 − x2) (x1 + x2) 16 + (Y1 − Y2) (Y1 + Y2) 4 = 0. From the midpoint coordinate formula, we can get 12 (x1 + x2) = 2, 12 (Y1 + Y2) = 1, KAB = Y1 − y2x1 − x2 = - X1 + x24 (Y1 + Y2) = 12
If the ellipse X & # 178 / 16 + Y & # 178 / 4 = 1 passes through point m (1,1), and the linear equation of the chord bisected by this point is
Let the ends of strings be a (x1, Y1), B (X2, Y2)
∴ x1+x2=2,y1+y2=2
∵ a, B on the ellipse X & # 178 / 16 + Y & # 178 / 4 = 1,
That is, in the ellipse X & # 178; + 4Y & # 178; = 16
∴ x1²+4y1²=16 ①
x2²+4y2²=16 ②
①-②
∴ (x1-x2)(x1+x2)+4(y1-y2)(y1+y2)=0
∴ 2(x1-x2)+8(y1-y2)=0
∴ k(AB)=(y1-y2)/(x1-x2)=-2/8=-1/4
The linear equation is Y-1 = (- 1 / 4) (x-1)
That is, x + 4y-5 = 0
The linear equation of the chord with the point m (1,1) in the ellipse x216 + y24 = 1 as the midpoint is___ ．
Let m (1,1) be the point where the chord at the middle point intersects the ellipse at a (x1, Y1), B (X2, Y22). Then x2116 + y214 = 1, x2216 + y224 = 1, subtracting (x1 + x2) (x1-x2) 16 + (Y1 + Y2) (y1-y22) 4 = 0. ∵ 1 = X1 + X22, 1 = Y1 + Y22, KAB = y1-y2x1-x2.. ∵ 216 + 2kab4 = 0, and solving KAB = - 14
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