Through the focus of the ellipse X225 + Y29 = 1, the length of the chord AB with an inclination angle of 45 ° is______ .

Through the focus of the ellipse X225 + Y29 = 1, the length of the chord AB with an inclination angle of 45 ° is______ .

∵ the elliptic equation is X225 + Y29 = 1, and the focal points are F1 (- 4, 0), F2 (4, 0), respectively. Let the equation of the straight line AB be y = x + 4. By eliminating y from ab equation and elliptic equation, we can get 34x2 + 200X + 175 = 0. Let a (x1, Y1), B (X2, Y2), we can get X1 + x2 = - 10017, x1x2 = 17534. Therefore, | ab | = 2 ·| x1-x2 | = 9017
If the chord ratio of the ellipse is known to be abx21 + 2K, then the focus of the ellipse is 2K + 2K______ .
According to the definition of ellipse, we know that: laf1l + laf2l = lbf1l + lbf2l = 2K + 2, and the circumference of triangle abf2 = labl + laf2l + lbf2l = laf1l + laf2l + lbf1l + lbf2l = 4K + 2 = 8, we get K + 2 = 4, k = 2 ∧ a = K + 2 = 2, e = CA = 12, so the answer is: 1
If a line passing through a focus F1 of ellipse 4x2 + 2Y2 = 1 intersects with ellipse at two points a and B, then a and B and another focus F2 of ellipse form △ abf2, then the perimeter of △ abf2 is ()
A. 2B. 22C. 2D. 1
Ellipse 4x2 + 2y22 = 1, that is & nbsp; x214 + y212 = & nbsp; 1, a = 22, B = 12, C = 12. The circumference of △ abf2 is (| AF1 | + | af2 | + (| BF1 | + | BF2 |) = 2A + 2A = 4A = 22, so B
If the hyperbola x29k2 − y24k2 = 1 and the circle x2 + y2 = 1 have no common point, then the value range of the real number k is______ .
∵ hyperbola x29k2 − y24k2 = 1 and circle x2 + y2 = 1 have no common point, ∵ 3K | 1, ∵ K | 13. The solution is k | 13 or K | 13. The value range of real number k is {K | K | 13 or K | 13}. The answer is {K | K | 13 or K | 13}
Given sin θ = 4 / 5 and cos θ = - 3 / 5, which quadrant does angle 2 θ belong to?
sin2θ=2sinθcosθ=2*(4/5)*(-3/5)=-24/25
cos2θ=cosθ^2-sinθ^2=-7/25
SO 2 theta is in the third quadrant
If the point (- 2,2) is m on the image of function y = 1 / 2x + 2m
-1+2m=2 m=1.5
Hyperbola x ^ 2 / (9K ^ 2) - y ^ / (4K ^ 2) and circle x ^ 2 + y ^ 2 = 1 have no common point. Find the range of K
As long as the vertex of the hyperbola is outside the unit circle, then 3 k > 1,
k1/3
Given cos (π + a) = - 1 / 2, calculate (1). Sin (2 π - a) (2) sin (a + (2n + 1) π) + a - (2n + 1) π
Calculate (1). Sin (2 π - a) (2) {sin [a + (2n + 1) π] + [Sina - (2n + 1) π]} / sin (a + 2n π) × cos (a-2n π)
cos（π+a）=-cosa=-1/2
cosa=1/2
sin²a+cos²a=1
sina=±√3/2
So. Sin (2 π - a)
=-sina
=±√3/2
｛sin[a+(2n+1）π]+[sina-(2n+1）π]｝/sin（a+2nπ）×cos（a-2nπ）
=(-sina-sina)/(sinacosa)
=-2tana
=±2√3
cos(π+a)=cos(π)cos(a)-sin(π)sin(a) = -cosa = -1/2.
So, cos (a) = 1 / 2;
sin(2π-a)=sin(-a)=-sin(a)=±√3/2;
｛sin[a+(2n+1）π]+[sina-(2n+1）π]｝/sin（a+2nπ）×cos（a-2nπ）
=[sin(a+π)+sin(a-π)]/sin(a)×cos(a)
=[-sin(a)-sin(a)]/sin(a)×cos(a)
=-2×cos(a)=-2
If the point (m, n) is on the image of function y = 2x = 1, then the value of 2m-n is the process of solving
Is your problem wrong? Is it 2x + 1 or 2x-1? Since the point is on the line, it means that the abscissa and ordinate of the point are substituted into the expression. The equation holds. Take + 1 as an example to substitute n = 2m + 1 and move to 2m-n = - 1
It is known that the vertex of the parabola is at the origin, the axis of symmetry is the X axis, and the focus is on the curve X ^ 2 / 4-y ^ 2 / 2 = 1
The focus of a parabola can only be on the X or Y axis,
The focus is the vertex of the curve X ^ 2 / 4-y ^ 2 / 2 = 1
Hyperbola with (0,2) and (0,2)
(1) The vertex is (2,0), and the parabolic equation y & # 178; = 8x
(2) The vertex is (- 2,0), and the parabolic equation y & # 178; = - 8x