answer:1. **Setting Up the Problem**: - Given the ellipse ( E: frac{x^2}{3} + y^2 = 1 ), with left focus ( F ). - A line ( l ) with positive slope passes through ( F ) and intersects the ellipse at points ( A ) and ( B ). - At points ( A ) and ( B ), perpendiculars are drawn to line ( l ), intersecting the ( x )-axis at points ( M ) and ( N ), respectively. - We need to find the minimum value of ( |M N| ). 2. **Equation of Line ( AB )**: - Suppose the equation of line ( AB ) is ( x = ny - sqrt{2} ) and intersects the ellipse at points ( A left( x_1, y_1 right) ) and ( B left( x_2, y_2 right) ). 3. **Finding Points ( M ) and ( N )**: - Line ( AM ) is perpendicular to line ( l ), hence its equation is ( y - y_1 = -n (x - x_1) ). - Setting ( y = 0 ) (since ( M ) and ( N ) lie on the ( x )-axis): [ 0 - y_1 = -n (x_M - x_1) implies x_M = x_1 + frac{y_1}{n} ] - Similarly, for point ( N ): [ x_N = x_2 + frac{y_2}{n} ] 4. **Distance ( |M N| )**: - Therefore, [ x_M - x_N = left( x_1 + frac{y_1}{n} right) - left( x_2 + frac{y_2}{n} right) = (x_1 - x_2) + frac{y_1 - y_2}{n} ] - Hence, [ |M N| = left| left( n + frac{1}{n} right)(y_1 - y_2) right| ] - Using the quadratic system: [ begin{cases} x = ny - sqrt{2} frac{x^2}{3} + y^2 = 1 end{cases} Rightarrow (n^2 + 3)y^2 - 2sqrt{2}ny - 1 = 0 ] 5. **Roots of the Quadratic Equation**: - Solving the quadratic equation for ( y ): [ y_1, y_2 = frac{ sqrt{2}n pm sqrt{ (2n^2 + 1) } }{ n^2 + 3 } ] - Distance ( y_1 - y_2 ): [ y_1 - y_2 = frac{ 2 sqrt{2}n }{ n^2 + 3 } ] 6. **Substituting into Distance Formula**: - Substituting ( y_1 - y_2 ) into ( |M N| ): [ |M N| = left( n + frac{1}{n} right) left| frac{2 sqrt{2}n}{n^2 + 3} right| = frac{2 sqrt{2}(n^2 + 1)}{n(n^2 + 3)} ] - Simplifying: [ frac{n^2 + 1}{n} cdot frac{2 sqrt{2}n}{n^2 + 3} = frac{2 sqrt{2} (n^2 + 1)}{n^2 + 3} ] 7. **Further Simplification**: - Set ( t = sqrt{1 + frac{1}{n^2}} ) such that ( t > 1 ): [ frac{2 sqrt{3} t^3}{3 t^2 - 2} ] - Let ( u = frac{1}{t} ): [ frac{2 sqrt{3}}{3 u - 2 u^3} ] 8. **Minimizing the Function**: - Consider ( varphi(u) = 3u - 2u^3 ): [ varphi'(u) = 3 - 6u^2 Rightarrow varphi(u) text{ is increasing on } (0, frac{sqrt{2}}{2}) text{ and decreasing on } (frac{sqrt{2}}{2}, 1) ] - Thus, ( varphi(u)_{max} = varphileft( frac{sqrt{2}}{2} right) = sqrt{2} ) 9. **Conclusion**: - Hence, the minimum value of ( |M N| ) is ( sqrt{6} ), which occurs when ( n = 1 ). [ boxed{sqrt{6}} ]

answer:Both lines 3x - 4y = 40 and 3x - 4y = 0 are parallel. The line equidistant between them is found by averaging the constants: [ frac{40 + 0}{2} = 20. ] So, the equidistant line is 3x - 4y = 20. Next, we solve the system of equations: 1. 3x - 4y = 20 2. x - 2y = 0 (from the line on which the center lies). From the second equation, x = 2y. Substituting into the first equation: [ 3(2y) - 4y = 20 ] [ 6y - 4y = 20 ] [ 2y = 20 ] [ y = 10. ] Then, using x = 2y: [ x = 2 times 10 = 20. ] Therefore, the center of the circle is boxed{(20, 10)}.

answer:The problem asks for the coefficient of x^3y^5 in the expanded form of (x+y)^8. According to the Binomial Theorem, the expansion of (x+y)^n includes terms of the form binom{n}{k}x^k y^{n-k} where k can range from 0 to n. For the x^3y^5 term: 1. We need k=3 so that the term has x^3. 2. With k=3, the corresponding term in the expansion is binom{8}{3}x^3y^{8-3} which reduces to binom{8}{3}x^3y^5. We calculate binom{8}{3} as: [ binom{8}{3} = frac{8 times 7 times 6}{3 times 2 times 1} = 56 ] Thus, the coefficient of the x^3y^5 term in the expansion of (x+y)^8 is boxed{56}.

answer:If (x = 0,) then the expression is equal to 0. Assuming (x neq 0,) we can simplify the expression by dividing the numerator and denominator by (x^4,) yielding: [ frac{1}{x^4 + 2x^2 + 4 + frac{8}{x^2} + frac{16}{x^4}}. ] Using the AM-GM inequality: [ x^4 + frac{16}{x^4} geq 2 sqrt{x^4 cdot frac{16}{x^4}} = 8, ] [ 2x^2 + frac{8}{x^2} geq 2 sqrt{2x^2 cdot frac{8}{x^2}} = 8. ] Thus, the denominator is at least: [ x^4 + 2x^2 + 4 + frac{8}{x^2} + frac{16}{x^4} geq 8 + 8 + 4 = 20. ] Therefore, the fraction becomes: [ frac{1}{x^4 + 2x^2 + 4 + frac{8}{x^2} + frac{16}{x^4}} leq frac{1}{20}. ] Equality occurs when (x^4 = frac{16}{x^4}) and (2x^2 = frac{8}{x^2},) solving these gives (x = sqrt[4]{16} = 2.) Thus, the maximum value is (boxed{frac{1}{20}}.)