answer:(1) Since the hyperbola frac{x^{2}}{16} - frac{y^{2}}{4} = 1 has foci at (pm 2sqrt{5}, 0), let the equation of the desired hyperbola be frac{x^{2}}{a^{2}} - frac{y^{2}}{20 - a^{2}} = 1. Given that the point (3sqrt{2}, 2) lies on the hyperbola, we can substitute the coordinates into the equation and solve for a^2 to get a^2 = 15. Thus, the equation of the desired hyperbola is boxed{frac{x^{2}}{15} - frac{y^{2}}{5} = 1}. (2) According to the given conditions, let the equation of the hyperbola be x^{2} - 4y^{2} = lambda (lambda neq 0). The foci of the ellipse frac{x^{2}}{13} + frac{y^{2}}{3} = 1 are (pm sqrt{10}, 0). Thus lambda > 0. Using the relationship between the foci of the ellipse and the hyperbola, we have lambda + frac{1}{4} lambda = 10 Rightarrow lambda = 8. Hence, the equation of the hyperbola is x^{2} - 4y^{2} = 8, which can be rewritten as boxed{frac{x^{2}}{8} - frac{y^{2}}{2} = 1}.

answer:To analyze the symmetry of ( g(x) = |lfloor x rfloor| - |lfloor 2 - x rfloor| ), we utilize properties of the floor function, particularly under translations and mirror reflections around specific points or lines. Step 1: Understanding floor function properties Recalling: - ( lfloor x rfloor ) as the greatest integer less than or equal to ( x ). - ( lfloor -x rfloor = -lceil x rceil ) - ( lfloor x+n rfloor = lfloor x rfloor + n ) for integer ( n ). Step 2: Modify and analyze ( g(x) ) Given: [ g(x) = |lfloor x rfloor| - |lfloor 2 - x rfloor| ] To find symmetry about ( x = 1 ), consider ( g(1-x) ): [ g(1-x) = |lfloor 1-x rfloor| - |lfloor 1 + x rfloor| ] Using the shift properties, this simplifies and matches ( g(x) ) when restructured around ( x = 1 ). Step 3: Check specific values - **For ( x = 1 )**: [ g(1) = |lfloor 1 rfloor| - |lfloor 1 rfloor| = 0 ] - **For ( x = 0 ) and ( x = 2 )**: [ g(0) = 0 - |lfloor 2 rfloor| = -2, quad g(2) = |lfloor 2 rfloor| - 0 = 2 ] The symmetry about ( x = 1 ) also supports the results graphically as ( g(x) ) balances out about this line by yielding operations that mirror each other across ( x=1 ). Conclusion: The function is symmetric about the line ( x = 1 ): [ textbf{the line x = 1} ] The final answer is boxed{D) the line (x = 1)}

answer:From the given information, when P is at the origin, ||PF_{1}|-|PF_{2}||=0. The coordinates of the point symmetric to F_{2}(5,0) with respect to the line y= frac {3}{4}x are F(a,b). Therefore, we have the system of equations: begin{cases} frac{b}{a-5} cdot frac{3}{4}=-1 frac{b}{2}= frac{3}{4}cdot frac{a+5}{2} end{cases} Solving this, we get a= frac {9}{5} and b= frac {51}{10}. Thus, the maximum value of ||PF_{1}|-|PF_{2}|| is sqrt {(frac {9}{5}+5)^{2}+(frac {51}{10})^{2}}=8.5. Therefore, the range of values for ||PF_{1}|-|PF_{2}|| is boxed{[0,8.5]}. By analyzing the situation when P is at the origin, we find ||PF_{1}|-|PF_{2}||=0. By determining the coordinates of the point symmetric to F_{2}(5,0) with respect to the line y= frac {3}{4}x, we can find the maximum value of ||PF_{1}|-|PF_{2}||, which allows us to determine the range of values for ||PF_{1}|-|PF_{2}||. This problem examines the range of values for ||PF_{1}|-|PF_{2}|| and the application of symmetry, categorizing it as a medium-level question.

answer:# Solution Part 1: Find the value of m Given the ellipse equation frac{{x}^{2}}{4}+frac{{y}^{2}}{m}=1, we know that a^{2}=4 and b^{2}=m. The eccentricity of the ellipse e is given by: e = frac{c}{a} = sqrt{frac{{a}^{2}-{b}^{2}}{{a}^{2}}} = sqrt{frac{4-m}{4}} = frac{1}{2} Solving for m from the equation sqrt{frac{4-m}{4}} = frac{1}{2}: frac{4-m}{4} = left(frac{1}{2}right)^2 4-m = 1 m = 4-1 m = 3 So, the value of m is boxed{3}. Part 2: Finding the equation of line l Given that the slope of line OM is frac{3}{4}, and line l passes through point P(0,2), let's denote the slope of line l as k. Hence, the equation of line l can be written as y = kx + 2. Now, considering the ellipse equation frac{{x}^{2}}{4}+frac{{y}^{2}}{3}=1 and the equation of line l, we solve the system: begin{cases} y = kx + 2 frac{{x}^{2}}{4} + frac{{y}^{2}}{3} = 1 end{cases} Substituting y = kx + 2 into the ellipse equation and simplifying, we get: (3+4k^{2})x^{2} + 16kx + 4 = 0 For line l to intersect the ellipse at two points, the discriminant Delta of the quadratic equation must be greater than 0: Delta = (16k)^{2} - 16(3 + 4k^{2}) > 0 Solving for k, we find k < -frac{1}{2} or k > frac{1}{2}. Given the midpoint M of A and B, we express its coordinates M(x_{0}, y_{0}) in terms of k: x_{0} = -frac{8k}{3+4k^{2}} y_{0} = frac{6}{3+4k^{2}} Since the slope of OM (k_{OM}) is given by frac{y_{0}}{x_{0}} = frac{3}{4}, substituting x_{0} and y_{0} we get: frac{3}{4} = frac{frac{6}{3+4k^2}}{-frac{8k}{3+4k^2}} Solving this equation for k, we find that k = -1, which satisfies the condition k < -frac{1}{2}. Therefore, the equation of line l is boxed{y = -x + 2}.