answer:This problem primarily involves a comprehensive understanding of conic sections. The circle {C}_{1}:{x}^{2}+{(y-2)}^{2}=4 passes through the origin (0,0), and the parabola {C}_{2}:{y}^{2}=2px (p > 0) also passes through the origin (0,0). Therefore, the line y=kx (k > 0) intersects the circle and the parabola at points A and B. The distance d from the circle's center to the line is d=sqrt{{2}^{2}-{left( frac{4 sqrt{5}}{5}right)}^{2}}=frac{2 sqrt{5}}{5}=frac{left|-2right|}{ sqrt{{k}^{2}+1}}. Solving for k, we get k=2 or k=-2 (discard the negative value). So, we have the system of equations: begin{cases} y=2x {y}^{2}=2px end{cases}. Solving this system of equations yields: begin{cases} x=frac{p}{2} y=p end{cases}. Now, left|ABright|=sqrt{{left( frac{p}{2}right)}^{2}+{p}^{2}}=frac{8 sqrt{5}}{5}. Solving for p, we get p=frac{16}{5}. Therefore, the equation of the parabola {C}_{2} is: boxed{{y}^{2}=frac{32}{5}x}.

answer:Let's denote the speed of contestant A as 3x and the speed of contestant B as 4x, where x is a common multiplier for their speeds. Since A has a start of 200 m, when B starts the race, A is already 200 m ahead. We need to find out how long it takes for B to cover the 500 m of the race and then see how far A has gone in that same time. The time it takes for B to finish the race is the distance B runs divided by B's speed: Time for B = Distance / Speed = 500 m / (4x) During this time, A is also running. The distance A covers in the same time is A's speed multiplied by the time for B: Distance A covers = Speed of A * Time for B = (3x) * (500 m / (4x)) We can simplify this by canceling out the x: Distance A covers = (3/4) * 500 m = 375 m Now, we need to add the 200 m head start that A had: Total distance A covers = 375 m + 200 m = 575 m Since the race is only 500 m long, we need to find out how much of the 575 m is beyond the finish line. This will tell us by how many meters A wins the race: A's winning distance = Total distance A covers - Race distance = 575 m - 500 m = 75 m Therefore, A wins the race by boxed{75} meters.

answer:1. **Suppose the statement is true:** - Assume that it is indeed possible for each pair of robbers to cross the river exactly once together. 2. **Denote the trips of the boat:** - Let the boat makes trips to cross robbers from the left bank to the right bank. The boat is two-seater, so it can carry either one or two robbers at a time. 3. **Calculate the number of times the boat is used for pairs:** - There are 40 robbers, and each pair of robbers crosses the river together exactly once. The number of distinct pairs that can be formed from 40 robbers is computed using the combination formula: [ binom{40}{2} = frac{40 cdot 39}{2} = 780 ] - Therefore, the boat must have made 780 trips carrying pairs. 4. **Determine the total trips:** - Since 780 trips carry pairs, the total number of trips must also include some odd number of trips carrying single robbers to return the boat back to left shore if we're considering all possible trips including returning. 5. **Counting odd-numbered single trips:** - To cross all robbers, including returning for odd-numbered trips, consider: - Each robber needs to be on the right bank finally. - From combinatorics and partitioning, every robber pairing odd trips results the number cross-over consistency being even. - If any robber crosses an odd number of times alone in single, consistency leads to keeping it odd for calculation parity consistency otherwise it must be non-possible scenario. 6. **Analysis contradiction for odd trips assumptions:** - Suppose, however, someone from sequence crosses an unpair number odd times alone. Evidently sum parity causes total pairs+ odd-single-redunant trips to hold odd. Given number combinations and valid intersection anable each pairing computational wholeheartedly simplifies keeping everyone on final consistency to odd or even combinations where logically consistency preferred within constraints turns unfeasable. - Thus, restores back odd-condficits human couldn not redunantly missing trips odd like vice ensured matching return reaches counter contradiction. # Conclusion: Since it's impossible checking conditions across-intersections valid values consistently restored crossing possibilities back within following requisite constraints inherently impossible serving combinations on required exactly basis; [ boxed{ text{No, it is not possible.} } ]

answer:For the hyperbola frac{x^{2}}{a^{2}} - frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0), the distance from one focus to an asymptote is b. Given that frac{b}{2c} = frac{1}{4}, we have b = frac{1}{2}c and a = sqrt{c^{2} - b^{2}} = frac{sqrt{3}}{2}c. Thus, frac{b}{a} = frac{sqrt{3}}{3}. Hence, the equations of its asymptotes are x pm sqrt{3}y = 0. Therefore, the answer is boxed{text{C}}. From the problem, we know that frac{b}{2c} = frac{1}{4}, so b = frac{1}{2}c and a = sqrt{c^{2} - b^{2}} = frac{sqrt{3}}{2}c. Thus, frac{b}{a} = frac{sqrt{3}}{3}, from which we can find the equations of its asymptotes. This problem tests the properties and applications of hyperbolas. When solving, carefully read the problem and solve it meticulously.