Given the set a = {x | X & # 178; - 160}, find a ∪ B

Given the set a = {x | X & # 178; - 160}, find a ∪ B




The line y = x + 1 and ellipse 3x ^ 2 + y ^ 2 = 2 intersect at P and Q. The proof is that the circle with the diameter of the line PQ passes through the coordinate origin


By substituting y = x + 1 into 3x ^ 2 + y ^ 2 = 2, we get 3x ^ 2 + (x + 1) ^ 2 = 24x ^ 2 + 2x-1 = 0, XP * XQ = - 1 / 4; XP + XQ = - 2 / 4yp * YQ = (XP + 1) (XQ + 1) = XP * XQ + (XP + XQ) + 1 = - 1 / 4-2 / 4 + 1 = 1 / 4kpo * kqo = YP / XP * YQ / XQ = YP * YQ / (XP * XQ) = (1 / 4) / (- 1 / 4) = - 1, so Po is perpendicular to Qo, so o is in PQ



It is known that the right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is F2 (3,0) and the eccentricity is e. let the line y = KX intersect the ellipse at two points a and B, and m and N are line segments respectively
Af2, the midpoint of BF2. If the coordinate origin o is on the circle with Mn as the diameter, and √ 2 / 2 < e ≤ √ 3 / 2, find the value range of K


Maybe my answer seems to be more troublesome: it's a fractional line, and the multiplication sign is omitted. Let a coordinate be (x1, kx1), and B coordinate be (X2, kx2) - which is reflected in the straight line y = KX. Using the midpoint coordinate formula, we can know m coordinate ((x1 + 3) / 2, kx1 / 2), n coordinate ((x2 + 3) / 2, kx2 / 2) because the origin o is on the circle with Mn as the diameter, so om ⊥ on