answer:For an ellipse with a standard form frac{x^2}{a^2} + frac{y^2}{b^2} = 1, where a and b are the semi-major and semi-minor axes respectively (with a > b), the foci are located on the axis corresponding to the larger denominator. We're given that the foci are on the y-axis, which means that b^2 > a^2. In our given equation x^2 + ky^2 = 2, we can equate this to the standard form by dividing by 2: frac{x^2}{2} + frac{ky^2}{2} = 1 Comparing to the standard formula, we identify a^2 = 2 and b^2 = 2/k. Now, because we want b^2 > a^2, we have: frac{2}{k} > 2 Solving for k, we must flip the inequality sign when we multiply both sides by k/2 (since we're assuming k > 0): k < 1 Therefore, the range of k must be (0, 1), where k is a positive real number (since for the equation to represent an ellipse, k cannot be zero or negative, as this would result in no ellipse due to the negative square term or division by zero). So, the correct option is D: (0, 1). [ boxed{D: (0, 1)} ]

answer:To describe the set of points ((x, y)) in the plane such that (t^2 + yt + x ge 0) for all (t) with (-1 le t le 1), we need to analyze the quadratic expression (t^2 + yt + x). 1. **Evaluate the quadratic at the boundary points (t = -1) and (t = 1):** [ t = -1: quad (-1)^2 + y(-1) + x ge 0 implies 1 - y + x ge 0 implies x - y + 1 ge 0 implies x ge y - 1 ] [ t = 1: quad (1)^2 + y(1) + x ge 0 implies 1 + y + x ge 0 implies x + y + 1 ge 0 implies x ge -y - 1 ] 2. **Consider the minimum value of the quadratic (t^2 + yt + x):** The minimum value of a quadratic (at^2 + bt + c) occurs at (t = -frac{b}{2a}). For (t^2 + yt + x), the minimum occurs at: [ t = -frac{y}{2} ] We need to ensure that this minimum value is non-negative for (-1 le t le 1). Thus, we evaluate the quadratic at (t = -frac{y}{2}): [ left(-frac{y}{2}right)^2 + yleft(-frac{y}{2}right) + x ge 0 implies frac{y^2}{4} - frac{y^2}{2} + x ge 0 implies x ge frac{y^2}{4} ] 3. **Combine the conditions:** We have three conditions: [ x ge y - 1 ] [ x ge -y - 1 ] [ x ge frac{y^2}{4} ] 4. **Geometric interpretation:** - The inequality (x ge y - 1) represents the region to the right of the line (x = y - 1). - The inequality (x ge -y - 1) represents the region to the right of the line (x = -y - 1). - The inequality (x ge frac{y^2}{4}) represents the region to the right of the parabola (x = frac{y^2}{4}). The set of points ((x, y)) that satisfy all three conditions is the intersection of these regions. This region is bounded by the parabola (x = frac{y^2}{4}) for (y in [-2, 2]) and the lines (x = y - 1) and (x = -y - 1) for (|y| ge 2). The final answer is the region to the right of the parabola ( boxed{ x = frac{y^2}{4} } ) for (y in [-2, 2]) and the lines (x = y - 1) and (x = -y - 1) for (|y| ge 2).

answer:To find the constant ( k ), we need to expand the right side of the equation and then compare the coefficients of the corresponding terms on both sides of the equation. The given equation is: [ -x^2 - (k + 10)x - 8 = -(x - 2)(x - 4) ] First, let's expand the right side: [ -(x - 2)(x - 4) = -[x^2 - 4x - 2x + 8] ] [ -(x - 2)(x - 4) = -[x^2 - 6x + 8] ] [ -(x - 2)(x - 4) = -x^2 + 6x - 8 ] Now, let's compare the coefficients of the corresponding terms on both sides of the equation: The coefficient of ( x^2 ) on the left side is ( -1 ), and on the right side, it is also ( -1 ), so they match. The coefficient of ( x ) on the left side is ( -(k + 10) ), and on the right side, it is ( 6 ). So we have: [ -(k + 10) = 6 ] Now, let's solve for ( k ): [ -k - 10 = 6 ] [ -k = 6 + 10 ] [ -k = 16 ] [ k = -16 ] Therefore, the constant ( k ) is ( boxed{-16} ).

answer:Since the coordinates of the two endpoints of the line segment are (5,1) and (m,1), and this line segment is bisected by the line x-2y=0, the midpoint of the line segment, left(frac{5+m}{2}, 1right), lies on the line x-2y=0. Substituting the midpoint's coordinates into the equation of the line, we get frac{5+m}{2} - 2 = 0. Solving for m, we obtain m = -1. Thus, the final answer is boxed{-1}.