If the line L: y = KX + m (k is not equal to 0) intersects the ellipse C: x ^ / 4 + y ^ / 3 = 1 at two different points m and N, and the line segment Mn has a When the vertical bisector passes through the fixed point G (1 / 8, 0), the value range of K is obtained

If the line L: y = KX + m (k is not equal to 0) intersects the ellipse C: x ^ / 4 + y ^ / 3 = 1 at two different points m and N, and the line segment Mn has a When the vertical bisector passes through the fixed point G (1 / 8, 0), the value range of K is obtained

Let m point coordinate be (a, Ka + m) and N point coordinate be (B, KB + m), where the point is (C, KC + m) Mn, and the vertical bisector is Y - [KC + M] = (- 1 / k) (x-C) which passes through the fixed point G (1 / 8,0), so 0 - (KC + m) = (- 1 / k) (1 / 8-c) K & # 178; C + km = 1 / 8-c, C = (1 / 8-km) / (K & # 178; + 1) substitute y = KX + m into X -

The elliptic equation y ^ 2 / 9 + x ^ 2 = 1, if there is a line L intersecting with the ellipse e at two different points m and N, so that the line segment Mn is exactly bisected by the line x = - 1 / 2, try to find L
The range of inclination angle

Obviously, the slope of line L exists. Let the linear equation be y = KX + B and substitute it into the elliptic equation: (KX + b) ^ 2 + 9x ^ 2 = 9 (k ^ 2 + 9) x ^ 2 + 2kbx + B ^ 2-9 = 0x1 + x2 = - 2KB / (k ^ 2 + 9) = - 1B = (k ^ 2 + 9) / (2k) Δ = 4K ^ 2B ^ 2-4 (k ^ 2 + 9) (b ^ 2-9) = 36K ^ 2-36b ^ 2 + 324 > 0k ^ 2-B ^ 2 + 9 > 0

Given that the line L passes through the point P (- 1,2) and intersects the line segment with a (- 2, - 3), B (3,0) as the end point, the value range of the slope of the line L is obtained

The slope of the straight line AP is k = - 3 − 2 − 2 + 1 = 5, and the slope of the straight line BP is k = 0 − 23 + 1 = - 12. Let L and line AB intersect at point m, and M moves from a to B, and the slope becomes larger and larger. At a certain point, am will be parallel to the Y axis, and there is no slope. That is, K ≥ 5. After this point, the slope increases from - ∞ to - 12 of the straight line BP

Given that the line L passes through the point P (- 1,2) and intersects the line segment with a (- 2, - 3), B (3,0) as the end point, the value range of the slope of the line L is obtained

The slope of the straight line AP is k = - 3 − 2 − 2 + 1 = 5, and the slope of the straight line BP is k = 0 − 23 + 1 = - 12. Let L and line AB intersect at point m, and M moves from a to B, and the slope becomes larger and larger. At a certain point, am will be parallel to the Y axis, and there is no slope. That is, K ≥ 5. After this point, the slope increases from - ∞ to - 12 of the straight line BP