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The focal coordinates of the parabola y2 = - 8x are______ ．
∵ parabolic equation y2 = - 8x, the focus is on the x-axis, P = 4, the focus coordinate is (- 2,0), so the answer is (- 2,0)
The focus coordinate of parabola y = 2x2 is ()
A. （18，0）B. （0，18）C. （0，12）D. （12，0）
The standard equation of parabola y = 2x2 is x2 = 12Y, P = 14, the opening of parabola is upward, the focus is on the positive half axis of Y axis, so the focus coordinate is (0, 18 & nbsp;), so B
1. The focus coordinate of parabola y ^ 2 – x = 0 is___________ The Quasilinear equation is___________ 2. Find the elliptic equation with the same eccentricity as x ^ 2 / 5 + y ^ 2 / 4 = 1 and passing through the point (root 5,2)__________ . 3. The definition field of function y = root (5x – x ^ 2 + 6) is______________ 4. Given that the coordinates of point P (x, y) satisfy the following conditions: (1) x + y is less than or equal to 4; (2) y is greater than or equal to X; (3) x is greater than or equal to 1, and point O is the origin of the coordinates, then the minimum value of L Po L is equal to 1____________ , the maximum value is equal to__________________ .
1. (1 / 4,0); X = - 1 / 4.2. X ^ 2 / 10 + y ^ 2 / 8 = 1 or Y ^ 2 / 41 / 4 + x ^ 2 / 41 / 5 = 1.3. - 1 ≤ x ≤ 6.4. Radical 2; radical 10
The focal coordinates of the parabola y = 1 / AX2 are
x^2 = ay = 2py
p = a/2
Focus (0, a / 4)
The focus coordinate of parabola (Y-1) 2 (2 is square) = - 16 (x + 2) is? Answer (- 6,1)
(Y-1) 2 = - 16 (x + 2) is to shift the whole image of y2 = - 2px, [here P = 8, focus coordinate (- 4,0)] by 2 units to the left, and by 1 unit to the up, the focus coordinate becomes (- 6,1)
The focus coordinate of parabola y = x ^ 2-2x is that the answer is (1, - 3 / 4), which requires process and answer knowledge
Y = x ^ 2-2x = (x-1) ^ 2-1, (x-1) ^ 2 = y + 1 is obtained by moving x ^ 2 = y one unit to the right and one unit down,
So the focus coordinate has the same translation as (0,1 / 4), which is (1, - 3 / 4)
It is known that the center of the hyperbola is the coordinate origin, the focus is on the coordinate axis, the focal length is 10, and through the point P (0,40), the equation of the hyperbola is solved
If the focal length 2C = 10, C = 5, the center is the origin of the coordinate, the focus is on the coordinate axis and passes through the point (0,4), then the focus is on the Y axis, a = 4, a ^ 2 = 16
So B ^ 2 = 5 ^ 2-4 ^ 2 = 9, B = 3
So the equation is: y ^ 2 / 16-x ^ 2 / 9 = 1
If the chord is perpendicular to the major axis through a focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), then the length of the chord is?
Through the right focus F2 of the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1, make the chord perpendicular to the X axis, and calculate the chord length
c^2=25-16
c=3
F2(3,0)
So the string x = 3
Substituting
9/25+y^2/16=1
y^2-256/25=0
y1+y2=0,y1y2=-256/25
(y1-y2)^2=(y1+y2)^2-4y1y2=1024/25
|y1-y2|=32/5
So chord length = 32 / 5
Let F1 and F2 be the left and right focus of hyperbola X / a-y / b = 1, respectively. If there is a point a on the hyperbola, the angle f1af2 = 90 degrees, and the absolute value of AF1 = 3 times the absolute value of af2, then the eccentricity of hyperbola is?
The absolute value of AF1 = 3 times the absolute value of af2, and the angle f1af2 = 90 degrees, so the absolute value of F1F2 = 10 times the absolute value of af2
2A = absolute value of AF1 minus absolute value of af2 = 2 times absolute value of af2, 2C = absolute value of root sign 10 times absolute value of af2
Eccentricity = C / a = root 10 / 2
RELATED INFORMATIONS
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- 2. The edge length of rectangle ABCD is ab = 2, BC = 3, point P is a point on the edge of AD, q is any point on the edge of BC, connecting AQ, DQ, passing through point P as PE parallel DQ intersecting AQ at point E, PE parallel AQ intersecting DQ at point F. let the length of AP be x, try to find the functional relationship between the area of triangle PEF and X, and find out where the point P is, and what is the maximum area of triangle PEF? When the point q is where, the perimeter of the triangle ADQ is the smallest? (the process or method to determine where q is should be given, without proof)
- 3. It is proved that the quadrilateral ABCD is a parallelogram
- 4. If the circumference of the parallelogram ABCD is 50 cm and ab is 3 cm shorter than CD, then CD,
- 5. In rectangle ABCD, e is the midpoint of BC, and AE ⊥ ed, if AE = 2, root 2, calculate the area of rectangle
- 6. As shown in the figure, BD is the diagonal of parallelogram ABCD, AE ⊥ BD is in E, CF ⊥ BD is in F
- 7. The base of a parallelogram is 8 cm, the height is half of the base, and the area of a triangle equal to its base and height
- 8. As shown in figure 1-4-11, it is known that the perimeter of the quadrilateral ABCD is 25cm, and the distance between the opposite sides is de = 2cm and DF = 3cm respectively. Calculate the area (cm2) of the parallelogram
- 9. It is known that the circumference of a rectangle is 60cm, and the length is 20cm more than the width. Let the length of a rectangle be xcm and the width be YCM, and the binary linear equations about X and y are listed
- 10. In the triangle ABC, if a [- 1,2,3], B [2, - 2,3], C [0.5,2.5,3], then the length of the central line CD on the side of ab
- 11. Through the fixed point m (4,0), make a straight line L intersection parabola y ^ 2 = 4x at two points a and B, f is the focus of the parabola, and find the minimum value of △ Fab area?
- 12. The equation of circle C (x-1) ^ 2 + y ^ 2 = 9, point P is a moving point on the circle, the coordinate of fixed point a is (a, 0), and the vertical bisector of line AP and straight line CP intersect at point M (1) If a = - 1, find the trajectory equation of point M; (2) When a changes, try to discuss the type of locus of point M
- 13. From fixed point a (3,0) to fixed circle x ^ 2 + y ^ 2 = 2, make a straight line intersecting circle at point P, and find the trajectory equation of the midpoint of line AP
- 14. The fixed point a (3,0) is a point outside the circle x ^ 2 + y ^ 2 = 1, P is a moving point on the circle, the bisector of ∠ POA intersects Pa at Q, and the trajectory of Q point is obtained
- 15. Point Q moves on the curve X ^ 2 + y ^ 2 = 1, and the symmetric point of point Q about point a (1, - 1) is p. find the trajectory equation of P
- 16. Given that the two points in the Cartesian coordinate plane are a (2,2), B (- 1, - 2), P is on the X axis, and PA = Pb, the coordinates of point P are obtained
- 17. It is known that two points a (- 5,2) and B (- 1,7) in the rectangular coordinate plane can find point P on the coordinate axis, so that point PA = Pb
- 18. Given the points a (- 1,0), B (1,0), P is the moving point on the line 2x-y + 1 = 0. (1) when the vector PA * and the vector Pb take the minimum value, we find the coordinates of the op vector and Co And find the value of COS angle APB (2) if P satisfies PA + Pb = pa-pb, the coordinates of P
- 19. Given the points a (- 2,5) and B (2,4), find a point P on the straight line L: 2x-2y + 1 = 0, so that | PA | + | Pb |, and find the minimum value
- 20. Given the straight line L: 2x-y + 1 = 0 and the point a (- 1,2) B (0,3), try to find a point P on L, so that the value of PA = Pb is minimum, and find the minimum value