The value range of M is obtained when the line y = x + m intersects the ellipse x ^ 2 / 16 + y ^ 2 / 9 = 1 X square / 16 + y square / 9 = 1: 9x ^ 2 + 16y ^ 2-144 = 0 Take y = x + m into the elliptic equation, and 9x ^ 2 + 16 (x + m) ^ 2-144 = 0 Sorted: 25X ^ 2 + 32mx + 16m ^ 2-144 = 0 Because there are two intersections, so Δ > = 0 That is, Δ = (32m) ^ 2-4 * 25 * (16m ^ 2-144) > = 0 Sorted out: m ^ 2-25

The value range of M is obtained when the line y = x + m intersects the ellipse x ^ 2 / 16 + y ^ 2 / 9 = 1 X square / 16 + y square / 9 = 1: 9x ^ 2 + 16y ^ 2-144 = 0 Take y = x + m into the elliptic equation, and 9x ^ 2 + 16 (x + m) ^ 2-144 = 0 Sorted: 25X ^ 2 + 32mx + 16m ^ 2-144 = 0 Because there are two intersections, so Δ > = 0 That is, Δ = (32m) ^ 2-4 * 25 * (16m ^ 2-144) > = 0 Sorted out: m ^ 2-25

Δ = (32m) ^ 2-4 * 25 * (16m ^ 2-144) = 32 & sup2; M & sup2; - 25 * 64M + 25 * 64 * 9 > = 0, 16m & sup2; - 25m & sup2; + 25 * 9 > = 0 if 64 is deleted from both sides
M & sup2; - 25

X ^ 2 / 16 + y ^ 2 / 9 = 1, a and B are two points on the ellipse, and the vertical bisector of a and B intersects the X axis at P (x0,0), so the value range of x0 can be obtained
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Let a (x1, Y1), B (X2, Y2), the midpoint of line AB be point D, the coordinates of point d be ((x1 + x2) / 2, (Y1 + Y2) / 2), the slope of line AB be, (y2-y1) / (x2-x1), the slope of line DP be, k = - (x2-x1) / (y2-y1) and the equation of line DP be y - (Y1 + Y2) / 2 = - [(x2-x1) / (y2-y1)] * (x - (x1 + x2) / 2) 0 -

Let the focus of the ellipse x24 + y2 = 1 be points F1 and F2, and point p be a moving point on the ellipse. When ∠ f1pf2 is an obtuse angle, the value range of abscissa of point P is obtained

Let P (x, y), then F1 (- 3, 0), F2 (3, 0), and ∠ f1pf2 be the obtuse angle ⇔ pf21 + PF22 & lt; f1f22 ⇔ (x + 3) 2 + Y2 + (x-3) 2 + Y2 & lt; 12 ⇔ x2 + 3 + Y2 & lt; 6 ⇔ x2 + (1-x24) & lt; 3 ⇔ x2 & lt; 83 ⇔ - 263 & lt; X & lt; 263

Given that the focus of the ellipse x 29 + y 24 = 1 is F 1 and F 2, the coordinates of the upper moving point P of the ellipse are (XP, YP), and ∠ f 1pf 2 is an obtuse angle, the value range of XP is obtained

The focus of the ellipse X29 + y24 = 1 is F1 (− 5,0), F2 (5,0) (2 points) so, Pf1 = (- 5 − XP, − YP), PF2 = (5 − XP, − YP). And ∠ f1pf2 is an obtuse angle, so Pf1 · PF2 < 0, that is (− 5 − XP) (5 − XP) + y2p < 0 So, x2p − 5 + 4 − 49x2p ＜ 0, the solution is − 355 ＜ XP ＜ 355 (9 points) so the abscissa of point P is (− 355355 355). & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp (10 points)