answer:To approach this problem, let's start with understanding the given geometric configuration. We have a plane that intersects a cone in a parabola and a sphere in a circle. The key piece of information here is that the plane contains the axis of the cone. Consider the cone first. The fact that the plane contains the axis of the cone means that the intersection of the cone and the plane is a parabola that is symmetric about the axis of the cone. The vertex of this parabola must lie on the axis of the cone because the parabola is symmetric about it. Now, let's consider the sphere. The intersection of the sphere and the plane is a circle. This circle is also symmetric about the axis of the cone since the plane contains this axis. This symmetry implies that the center of the circle lies on the axis of the cone. The crucial insight is that both the vertex of the parabola and the center of the circle lie on the axis of the cone. Moreover, the point where the axis of the cone intersects the sphere is equidistant from every point on the circle (since it's the center of the sphere), and it also lies on the axis of the cone. Given these relationships, it follows that the vertex of the parabola, which lies on the axis of the cone, coincides with the point where the axis intersects the sphere. This intersection point is the center of the circle. Since the vertex of the parabola is also equidistant from every point on the circle and lies on the axis of the cone, it must be this center point, which inherently lies on the circle itself. Therefore, we conclude that the vertex of the parabola indeed lies on the circle formed by the intersection of the plane and the sphere. This elegant relationship showcases the harmony between the geometric shapes involved, highlighting the intricate connections that arise from their spatial relationships.

answer:To understand the relationship between the points on the parabola where the tangent line is parallel to the x-axis and the focus, we need to consider the slope of the tangent line to the parabola. The slope of the tangent line to the parabola y = x^2 at any point (x, x^2) is given by the derivative of the function with respect to x, which is 2x. For the tangent line to be parallel to the x-axis, its slope must be zero. Therefore, we need to find the points on the parabola where the slope of the tangent line is zero. Setting the slope 2x equal to zero gives us x = 0. This corresponds to the point (0, 0) on the parabola. However, we should note that a parabola has only one point where the tangent is parallel to the x-axis if we're looking for a horizontal tangent line that isn't at the vertex of a parabola given by y = a(x - h)^2 + k, a parabola of the form y = x^2 does indeed have its horizontal tangent at the vertex. Now, let's recall the definition of a parabola as the set of all points that are equidistant from the focus and the directrix. The focus of the parabola y = x^2 is at (0, 1/4), and the directrix is the horizontal line y = -1/4. The point (0, 0) on the parabola is indeed equidistant from the focus and the directrix, as its distance from both is 1/4. However, a deeper relationship exists in the context of parabolas with a more general form like y = ax^2, where 'a' is not equal to 1. In that case, for any parabola, the points where the tangent is parallel to the x-axis lie on a line that is the same distance from the focus as from the directrix but is on the directrix's opposite side from the vertex of the parabola for parabolas with their vertex at (0,0), it being the x-axis, and the points come at x = 0 if that horizontal tangent is at the vertex itself. In this specific case, y = x^2, since there's only one point where the tangent line is parallel to the x-axis and it is at the vertex, the relationship holds true in a more straightforward manner: the point where the tangent is parallel to the x-axis, the vertex, is on the axis of symmetry that contains the focus and is equidistant from the focus and the directrix.

answer:To solve the equation z^4 = -4, we first rewrite it as z^4 + 4 = 0. This is a quartic equation, but we can simplify it by noticing that -4 can be written in polar form as 4e^(iπ). Let's express z in polar form as re^(iθ). Then, z^4 = r^4e^(i4θ). Equating this to 4e^(iπ), we get: r^4e^(i4θ) = 4e^(iπ) Since the magnitudes and angles of both sides must be equal, we have: r^4 = 4 => r = ±√2 (since r is a real number and we're taking the fourth root) And for the angles: 4θ = π + 2kπ (where k is an integer) θ = π/4 + kπ/2 (dividing both sides by 4) Now, let's consider the possible values of k: For k = 0, θ = π/4 For k = 1, θ = 3π/4 For k = 2, θ = 5π/4 For k = 3, θ = 7π/4 Any further values of k will give us angles that are equivalent to these four. So, we have four possible values for z: z = √2e^(iπ/4) = √2(cos(π/4) + i sin(π/4)) = √2(1/√2 + i/√2) = 1 + i z = √2e^(i3π/4) = √2(cos(3π/4) + i sin(3π/4)) = √2(-1/√2 + i/√2) = -1 + i z = √2e^(i5π/4) = √2(cos(5π/4) + i sin(5π/4)) = √2(-1/√2 - i/√2) = -1 - i z = √2e^(i7π/4) = √2(cos(7π/4) + i sin(7π/4)) = √2(1/√2 - i/√2) = 1 - i These are the four solutions to the equation z^4 = -4, expressed in the form a + bi.

answer:Here's a revised version of the sentence: The village was surrounded by a vast forest teeming with a diverse array of wildlife. I made a few changes to enhance clarity and create a more vivid image. By using surrounded instead of around, the sentence now clearly conveys the forest's relationship to the village. I also replaced big with vast to use a more descriptive adjective, and many animals with a diverse array of wildlife to add more depth and variety to the scene.