a ball is released at the point (d, h) on to the ground which has the shape of a hyperbola with equation −(x^2 /a^2 )+(((y+b)^2 )/b^2 )=1. the collision between the ball and the ground is elastic. describe the motion the ball will have. will the ball return back to this point after some time? if yes, after how many times collision with the ground will this happen?---**Chart Description:** * **Type:** Coordinate system diagram with a curve and labeled points/lines. * **Main Elements:** * **Coordinate Axes:** Horizontal x-axis pointing right, vertical y-axis pointing up. The origin is labeled 'O'. The y-axis is labeled 'y' and the x-axis is labeled 'x'. * **Curve:** A red curve opening upwards, passing through the origin O. It appears to be symmetric about the y-axis, resembling a parabola. * **Points:** * Point A is a blue dot located in the upper right quadrant. * Point P is a red label on the red curve in the right half of the plane. * **Lines:** * A dashed horizontal line extends from the y-axis to the horizontal position of point A. This line segment is labeled 'd'. * A solid blue vertical line extends downwards from point A to a point on the red curve. This line segment has a blue arrow pointing downwards, indicating a direction. The endpoint of this line on the curve is near the label 'p'. * **Labels:** * 'y' labels the vertical axis. * 'x' labels the horizontal axis. * 'O' labels the origin (intersection of x and y axes). * 'h' is labeled on the left side of the y-axis, indicating a vertical distance from the origin along the y-axis. Its endpoint is not explicitly marked or connected to anything specific in this diagram, but its position suggests it represents a height or y-coordinate. * 'd' is labeled above the dashed horizontal line, indicating a horizontal distance. Its position suggests it might represent the x-coordinate of point A. * 'A' labels the blue point in the upper right. * 'p' labels the point on the red curve where the vertical blue line from A terminates and where a curved blue arrow originates. * **Arrows:** * A blue arrow points vertically downwards from point A towards the red curve. * A curved blue arrow originates from the point on the red curve labeled 'p' and points leftwards, tangent to the curve, indicating a direction along the curve. **Textual Information:** * Labels present in the image: y, x, O, h, d, A, p. (Note: There is no question stem, options, or other explanatory text in the image provided, only the diagram and its labels.)

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We have a ball released from point (d, h) onto a hyperbolic ground surface. The hyperbola has equation minus x squared over a squared plus y plus b quantity squared over b squared equals 1. This hyperbola opens upward. When the ball collides with the ground, the collision is elastic, meaning the angle of incidence equals the angle of reflection relative to the surface tangent. Now let's analyze the first collision. The ball falls vertically from point (d, h) and hits the hyperbola at the collision point. We find this point by substituting x equals d into the hyperbola equation. The collision point has coordinates (d, y) where y equals the square root of b squared times one plus d squared over a squared, minus b. At the collision point, we calculate the tangent slope using calculus. The derivative dy dx equals bx over a squared times the square root of one plus x squared over a squared. For elastic collision, the angle of incidence equals the angle of reflection relative to this tangent line. Let's observe the trajectory pattern through multiple collisions. The ball starts at point (d, h) and follows a predictable path. After the first collision at P1 with coordinates (d, y1), the ball reflects and travels to P2 at (-d, y1) due to the hyperbola's symmetry. From P2, it bounces upward and then returns to P1. This creates a closed trajectory pattern. The hyperbola's symmetric properties ensure that each collision point can be calculated precisely using the reflection principle. Now let's investigate the periodicity of this motion. The key question is: will the ball return to its starting point (d, h)? The answer is yes! Due to the hyperbola's reflective symmetry properties, the ball follows a closed trajectory. After exactly 4 collisions with the ground, the ball returns to point (d, h). The first collision occurs at (d, y), the second at (-d, y), the third again at (-d, y), and the fourth back at (d, y). From there, the ball returns to its starting height. This creates a periodic motion with period equal to 4 collisions. Let's provide the mathematical proof for why the ball returns after exactly 4 collisions. The key lies in the hyperbola's symmetry properties. Since the hyperbola equation shows y equals f of x equals f of negative x, it's symmetric about the y-axis. Combined with the elastic collision law where incident angle equals reflected angle, this creates a predictable trajectory. Starting from (d,h), the ball hits point 1 at (d, y1), reflects to point 2 at (-d, y1), bounces to point 3 at (-d, y1), then returns to point 4 at (d, y1), and finally back to (d,h). This completes one cycle after exactly 4 collisions. This demonstrates the general principle that elastic collisions on conic sections often produce periodic motion due to their geometric properties.