answer：Okay, so I have this function f(x) = a^{x+2} + 1, and I need to find a fixed point that this function must pass through, regardless of the value of 'a', as long as a is greater than 0 and not equal to 1. Hmm, fixed points... I think that means a point (x, y) where no matter what 'a' is, when I plug in x into the function, I always get y. So, it's like a point that's consistent for all possible values of 'a'. Alright, let me think about this. The function is an exponential function, right? It has the form a raised to some power, plus 1. Exponential functions have certain properties. For example, when the exponent is 0, any non-zero number raised to the power of 0 is 1. That might be useful here. Looking at the exponent in the function, it's x + 2. If I can make that exponent equal to 0, then a^{0} would be 1, regardless of what 'a' is. So, if I set x + 2 = 0, then x would be -2. Let me try plugging x = -2 into the function to see what y is. f(-2) = a^{-2 + 2} + 1 = a^{0} + 1 = 1 + 1 = 2. Oh, so when x is -2, y is always 2, no matter what 'a' is. That seems like a fixed point because it doesn't depend on 'a'. Wait, is that the only fixed point? Let me check another value. Maybe if I set x + 2 to something else, like 1. Then x would be -1, and f(-1) = a^{1} + 1. But a^{1} is just 'a', and since 'a' can be any positive number except 1, this y-value would change depending on 'a'. So, that's not fixed. What if I set x + 2 to 2? Then x would be 0, and f(0) = a^{2} + 1. Again, this depends on 'a', so it's not fixed. It seems like the only way to get a fixed y-value is to make the exponent 0, which gives me a^{0} = 1, and then adding 1 gives me 2. So, the fixed point must be at (-2, 2). Let me visualize this. The graph of f(x) = a^{x+2} + 1 is an exponential curve. Since 'a' is positive and not equal to 1, it will either grow or decay exponentially. The "+1" at the end shifts the entire graph up by 1 unit. The term a^{x+2} can be rewritten as a^{x} * a^{2}, which is just scaling the exponential function by a factor of a^{2}. But regardless of how it's scaled or shifted, when x is -2, the exponent becomes 0, making a^{0} = 1, and then adding 1 gives 2. So, no matter how I scale or shift the graph, it will always pass through (-2, 2). Is there another way to think about this? Maybe by considering transformations of functions. The original function is f(x) = a^{x}. If I replace x with x + 2, it's a horizontal shift to the left by 2 units. Then, adding 1 shifts it up by 1 unit. So, the original function a^{x} passes through (0, 1), right? Because when x is 0, a^{0} is 1. After shifting left by 2, the point (0, 1) moves to (-2, 1). Then, shifting up by 1 unit moves it to (-2, 2). So, that confirms it again. The fixed point is (-2, 2). I think I'm confident now that (-2, 2) is the fixed point through which the function must pass, regardless of the value of 'a'. It's because when x is -2, the exponent becomes 0, making a^{0} = 1, and then adding 1 gives 2. This doesn't depend on 'a', so it's fixed.

answer：First, I need to determine the total number of minutes in 15 days. I'll start by converting days to hours and then hours to minutes.15 days multiplied by 24 hours per day equals 360 hours. Then, 360 hours multiplied by 60 minutes per hour equals 21,600 minutes.Next, I'll calculate the average megabytes per minute by dividing the total disk space by the total number of minutes. So, 25,500 MB divided by 21,600 minutes is approximately 1.1806 MB per minute.Finally, I'll round this value to the nearest tenth, which gives me 1.2 MB per minute.

answer：First, I need to determine the total number of students across all 6 classrooms. Since each classroom has 24 students, I multiply 24 by 6 to get the total number of students.Next, I calculate the total number of guinea pigs. Each classroom has 3 guinea pigs, so I multiply 3 by 6 to find the total number of guinea pigs.Finally, to find out how many more students there are than guinea pigs, I subtract the total number of guinea pigs from the total number of students.

answer：Okay, so I have this problem about a line and a parabola intersecting, and I need to find the standard equation of the parabola and then prove something about another line passing through a fixed point. Let me try to break this down step by step.First, part (1): I need to find the standard equation of the parabola C. The given parabola is ( y^2 = 2px ) where ( p > 0 ). The line given is ( l: y = k(x + frac{p}{2}) ). When ( k = frac{1}{2} ), the chord MN has a length of ( 4sqrt{15} ). Alright, so when ( k = frac{1}{2} ), the line becomes ( y = frac{1}{2}(x + frac{p}{2}) ). I need to find where this line intersects the parabola. Let me substitute ( y ) from the line into the parabola equation.So, substituting ( y = frac{1}{2}x + frac{p}{4} ) into ( y^2 = 2px ), I get:( left( frac{1}{2}x + frac{p}{4} right)^2 = 2px )Let me expand the left side:( left( frac{1}{2}x + frac{p}{4} right)^2 = left( frac{1}{2}x right)^2 + 2 cdot frac{1}{2}x cdot frac{p}{4} + left( frac{p}{4} right)^2 )Calculating each term:1. ( left( frac{1}{2}x right)^2 = frac{1}{4}x^2 )2. ( 2 cdot frac{1}{2}x cdot frac{p}{4} = frac{p}{4}x )3. ( left( frac{p}{4} right)^2 = frac{p^2}{16} )So, putting it all together:( frac{1}{4}x^2 + frac{p}{4}x + frac{p^2}{16} = 2px )Now, let's move all terms to one side to form a quadratic equation:( frac{1}{4}x^2 + frac{p}{4}x + frac{p^2}{16} - 2px = 0 )Combine like terms:First, let's get all terms with ( x^2 ), ( x ), and constants:- ( x^2 ) term: ( frac{1}{4}x^2 )- ( x ) terms: ( frac{p}{4}x - 2px = left( frac{p}{4} - 2p right)x = left( frac{p - 8p}{4} right)x = left( -frac{7p}{4} right)x )- Constant term: ( frac{p^2}{16} )So, the quadratic equation is:( frac{1}{4}x^2 - frac{7p}{4}x + frac{p^2}{16} = 0 )To make it easier, multiply all terms by 16 to eliminate denominators:( 4x^2 - 28px + p^2 = 0 )So, the quadratic equation is ( 4x^2 - 28px + p^2 = 0 ).Let me denote the roots as ( x_1 ) and ( x_2 ). From quadratic equation theory, we know that:( x_1 + x_2 = frac{28p}{4} = 7p )and( x_1x_2 = frac{p^2}{4} )Now, to find the length of chord MN, I need the distance between points M and N. The coordinates of M and N are ( (x_1, y_1) ) and ( (x_2, y_2) ) respectively.Since both points lie on the line ( y = frac{1}{2}x + frac{p}{4} ), their y-coordinates can be expressed in terms of their x-coordinates:( y_1 = frac{1}{2}x_1 + frac{p}{4} )( y_2 = frac{1}{2}x_2 + frac{p}{4} )So, the distance between M and N is:( |MN| = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} )Let me compute ( y_2 - y_1 ):( y_2 - y_1 = frac{1}{2}x_2 + frac{p}{4} - left( frac{1}{2}x_1 + frac{p}{4} right) = frac{1}{2}(x_2 - x_1) )So, the distance becomes:( |MN| = sqrt{(x_2 - x_1)^2 + left( frac{1}{2}(x_2 - x_1) right)^2} )Factor out ( (x_2 - x_1)^2 ):( |MN| = sqrt{(x_2 - x_1)^2 left( 1 + left( frac{1}{2} right)^2 right)} = |x_2 - x_1| sqrt{1 + frac{1}{4}} = |x_2 - x_1| sqrt{frac{5}{4}} = |x_2 - x_1| cdot frac{sqrt{5}}{2} )We are given that ( |MN| = 4sqrt{15} ). So,( |x_2 - x_1| cdot frac{sqrt{5}}{2} = 4sqrt{15} )Multiply both sides by 2:( |x_2 - x_1| sqrt{5} = 8sqrt{15} )Divide both sides by ( sqrt{5} ):( |x_2 - x_1| = 8sqrt{3} )Now, ( |x_2 - x_1| ) can be found using the formula ( |x_2 - x_1| = sqrt{(x_1 + x_2)^2 - 4x_1x_2} )We already have ( x_1 + x_2 = 7p ) and ( x_1x_2 = frac{p^2}{4} ). So,( |x_2 - x_1| = sqrt{(7p)^2 - 4 cdot frac{p^2}{4}} = sqrt{49p^2 - p^2} = sqrt{48p^2} = 4psqrt{3} )We found earlier that ( |x_2 - x_1| = 8sqrt{3} ). Therefore,( 4psqrt{3} = 8sqrt{3} )Divide both sides by ( 4sqrt{3} ):( p = 2 )So, the standard equation of the parabola is ( y^2 = 2px ) with ( p = 2 ), which gives ( y^2 = 4x ).Alright, that seems solid. Let me just recap to make sure I didn't make any mistakes. I substituted the line into the parabola, found the quadratic in x, used the sum and product of roots, calculated the distance between the two points, set it equal to the given length, solved for p, and got p=2. So, the parabola is ( y^2 = 4x ).Now, moving on to part (2). It says: The line passing through point M intersects the parabola at another point Q, and the line MQ passes through point B(1, -1). Prove that the line NQ passes through a fixed point.Hmm, okay. So, we have points M and N on the parabola, and another point Q such that line MQ passes through B(1, -1). We need to show that line NQ passes through a fixed point, regardless of where Q is, I suppose.First, let me note that from part (1), we have the parabola ( y^2 = 4x ). So, p=2, which means the focus is at (1,0), but maybe that's not directly relevant here.Given that the line l is ( y = k(x + 1) ) because ( p=2 ), so ( frac{p}{2} = 1 ). So, the line is ( y = k(x + 1) ). When k=1/2, we had points M and N. But in part (2), it's a general k or still k=1/2? Wait, the problem says "the line passing through point M intersects the parabola at another point Q". So, it's a different line, not necessarily the same line l. So, perhaps the line l is fixed as ( y = k(x + 1) ), but the line MQ is another line passing through M and Q, passing through B(1, -1).Wait, let me read it again: "The line passing through point M intersects the parabola at another point Q, and the line MQ passes through point B(1, -1)." So, the line MQ passes through B(1, -1). So, for any such Q, we need to show that NQ passes through a fixed point.Wait, but is Q arbitrary? Or is it determined by the condition that MQ passes through B? So, for each M, there is a specific Q such that MQ passes through B. Then, for each such Q, NQ passes through a fixed point. So, regardless of M and Q (as long as MQ passes through B), NQ passes through a fixed point.Alternatively, maybe it's for a particular M and N, but since the parabola is fixed, perhaps the fixed point is always the same.I think I need to parametrize points M and N, find Q such that MQ passes through B, then find the equation of NQ and show that it passes through a specific point regardless of the parameters.Let me try to parametrize the points on the parabola. For the parabola ( y^2 = 4x ), parametric coordinates can be written as ( (t^2, 2t) ). So, let me denote point M as ( (t_1^2, 2t_1) ) and point N as ( (t_2^2, 2t_2) ). Since M and N are intersection points of line l with the parabola, they correspond to parameters t1 and t2.Given that line l is ( y = k(x + 1) ), substituting into the parabola equation:( (k(x + 1))^2 = 4x )Which simplifies to:( k^2x^2 + 2k^2x + k^2 = 4x )Bringing all terms to one side:( k^2x^2 + (2k^2 - 4)x + k^2 = 0 )This quadratic in x has roots corresponding to points M and N. Let me denote the roots as x1 and x2, so:Sum of roots: ( x1 + x2 = frac{4 - 2k^2}{k^2} )Product of roots: ( x1x2 = 1 )Wait, but in parametric terms, the roots correspond to t1 and t2. For the parabola ( y^2 = 4x ), the parametric equations are ( x = t^2 ), ( y = 2t ). So, substituting into the line equation:( 2t = k(t^2 + 1) )Which gives:( kt^2 - 2t + k = 0 )So, this quadratic in t has roots t1 and t2. Therefore:Sum of roots: ( t1 + t2 = frac{2}{k} )Product of roots: ( t1t2 = 1 )That's an important relation: ( t1t2 = 1 ). So, if t1 is a parameter for M, then t2 is ( 1/t1 ). That might be useful.Now, point Q is another point on the parabola such that line MQ passes through B(1, -1). So, let me denote Q as ( (t_3^2, 2t_3) ). Then, the line MQ connects M ( (t1^2, 2t1) ) and Q ( (t3^2, 2t3) ). The equation of line MQ can be found using the two-point form.The slope of MQ is:( m = frac{2t3 - 2t1}{t3^2 - t1^2} = frac{2(t3 - t1)}{(t3 - t1)(t3 + t1)} = frac{2}{t3 + t1} )So, the slope is ( frac{2}{t3 + t1} ). Then, the equation of line MQ is:( y - 2t1 = frac{2}{t3 + t1}(x - t1^2) )We know that this line passes through B(1, -1). So, substituting x=1, y=-1 into the equation:( -1 - 2t1 = frac{2}{t3 + t1}(1 - t1^2) )Let me write this as:( -1 - 2t1 = frac{2(1 - t1^2)}{t3 + t1} )Multiply both sides by ( t3 + t1 ):( (-1 - 2t1)(t3 + t1) = 2(1 - t1^2) )Let me expand the left side:( -1 cdot t3 -1 cdot t1 - 2t1 cdot t3 - 2t1 cdot t1 = -t3 - t1 - 2t1t3 - 2t1^2 )So, the equation becomes:( -t3 - t1 - 2t1t3 - 2t1^2 = 2 - 2t1^2 )Let me bring all terms to the left side:( -t3 - t1 - 2t1t3 - 2t1^2 - 2 + 2t1^2 = 0 )Simplify:- The ( -2t1^2 ) and ( +2t1^2 ) cancel out.- So, we have:( -t3 - t1 - 2t1t3 - 2 = 0 )Let me factor terms:( -t3(1 + 2t1) - t1 - 2 = 0 )Let me solve for t3:( -t3(1 + 2t1) = t1 + 2 )Multiply both sides by -1:( t3(1 + 2t1) = -t1 - 2 )Therefore,( t3 = frac{ -t1 - 2 }{ 1 + 2t1 } )Simplify numerator and denominator:( t3 = frac{ - (t1 + 2) }{ 1 + 2t1 } )I can factor out a negative sign:( t3 = - frac{ t1 + 2 }{ 1 + 2t1 } )Alternatively, I can write this as:( t3 = frac{ t1 + 2 }{ -1 - 2t1 } = frac{ t1 + 2 }{ - (1 + 2t1) } = - frac{ t1 + 2 }{ 1 + 2t1 } )Either way, that's the expression for t3 in terms of t1.Now, since we have t3 expressed in terms of t1, we can find the coordinates of Q as ( (t3^2, 2t3) ).Now, we need to find the equation of line NQ and show that it passes through a fixed point.Point N is ( (t2^2, 2t2) ), and we know from earlier that ( t1t2 = 1 ), so ( t2 = frac{1}{t1} ).So, point N is ( left( left( frac{1}{t1} right)^2, 2 cdot frac{1}{t1} right) = left( frac{1}{t1^2}, frac{2}{t1} right) ).Point Q is ( (t3^2, 2t3) ), where ( t3 = - frac{ t1 + 2 }{ 1 + 2t1 } ).So, let me compute t3:( t3 = - frac{ t1 + 2 }{ 1 + 2t1 } )Let me compute t3^2:( t3^2 = left( - frac{ t1 + 2 }{ 1 + 2t1 } right)^2 = frac{ (t1 + 2)^2 }{ (1 + 2t1)^2 } )Similarly, 2t3 is:( 2t3 = 2 cdot left( - frac{ t1 + 2 }{ 1 + 2t1 } right) = - frac{ 2(t1 + 2) }{ 1 + 2t1 } )So, coordinates of Q are:( left( frac{ (t1 + 2)^2 }{ (1 + 2t1)^2 }, - frac{ 2(t1 + 2) }{ 1 + 2t1 } right) )Now, let's find the equation of line NQ. Points N and Q are:N: ( left( frac{1}{t1^2}, frac{2}{t1} right) )Q: ( left( frac{ (t1 + 2)^2 }{ (1 + 2t1)^2 }, - frac{ 2(t1 + 2) }{ 1 + 2t1 } right) )Let me denote point N as (xN, yN) and Q as (xQ, yQ).So,xN = ( frac{1}{t1^2} )yN = ( frac{2}{t1} )xQ = ( frac{ (t1 + 2)^2 }{ (1 + 2t1)^2 } )yQ = ( - frac{ 2(t1 + 2) }{ 1 + 2t1 } )Now, let's find the slope of NQ:( m_{NQ} = frac{ yQ - yN }{ xQ - xN } )Compute numerator:( yQ - yN = - frac{ 2(t1 + 2) }{ 1 + 2t1 } - frac{2}{t1} )Let me combine these terms. To do that, find a common denominator, which would be ( t1(1 + 2t1) ):( yQ - yN = - frac{ 2(t1 + 2) cdot t1 }{ t1(1 + 2t1) } - frac{ 2(1 + 2t1) }{ t1(1 + 2t1) } )Simplify numerator:( -2t1(t1 + 2) - 2(1 + 2t1) = -2t1^2 - 4t1 - 2 - 4t1 = -2t1^2 - 8t1 - 2 )So,( yQ - yN = frac{ -2t1^2 - 8t1 - 2 }{ t1(1 + 2t1) } )Now, compute denominator:( xQ - xN = frac{ (t1 + 2)^2 }{ (1 + 2t1)^2 } - frac{1}{t1^2} )Again, find a common denominator, which is ( t1^2(1 + 2t1)^2 ):( xQ - xN = frac{ t1^2(t1 + 2)^2 - (1 + 2t1)^2 }{ t1^2(1 + 2t1)^2 } )Let me expand the numerator:First term: ( t1^2(t1 + 2)^2 = t1^2(t1^2 + 4t1 + 4) = t1^4 + 4t1^3 + 4t1^2 )Second term: ( (1 + 2t1)^2 = 1 + 4t1 + 4t1^2 )So, numerator:( t1^4 + 4t1^3 + 4t1^2 - (1 + 4t1 + 4t1^2) = t1^4 + 4t1^3 + 4t1^2 - 1 - 4t1 - 4t1^2 )Simplify:- ( t1^4 )- ( +4t1^3 )- ( +4t1^2 - 4t1^2 = 0 )- ( -4t1 )- ( -1 )So, numerator is ( t1^4 + 4t1^3 - 4t1 - 1 )Therefore,( xQ - xN = frac{ t1^4 + 4t1^3 - 4t1 - 1 }{ t1^2(1 + 2t1)^2 } )Now, putting it all together, the slope ( m_{NQ} ) is:( m_{NQ} = frac{ -2t1^2 - 8t1 - 2 }{ t1(1 + 2t1) } div frac{ t1^4 + 4t1^3 - 4t1 - 1 }{ t1^2(1 + 2t1)^2 } )Dividing fractions is equivalent to multiplying by reciprocal:( m_{NQ} = frac{ -2t1^2 - 8t1 - 2 }{ t1(1 + 2t1) } times frac{ t1^2(1 + 2t1)^2 }{ t1^4 + 4t1^3 - 4t1 - 1 } )Simplify numerator and denominator:First, factor numerator of the first fraction:( -2t1^2 - 8t1 - 2 = -2(t1^2 + 4t1 + 1) )Denominator of the first fraction: ( t1(1 + 2t1) )Second fraction numerator: ( t1^2(1 + 2t1)^2 )Second fraction denominator: ( t1^4 + 4t1^3 - 4t1 - 1 )Let me see if the denominator ( t1^4 + 4t1^3 - 4t1 - 1 ) can be factored. Maybe factor by grouping:Group as ( (t1^4 + 4t1^3) + (-4t1 - 1) = t1^3(t1 + 4) - (4t1 + 1) ). Doesn't seem to factor nicely. Alternatively, maybe it's a quadratic in t1^2? Let me check:Let me denote ( u = t1^2 ), then the denominator becomes ( u^2 + 4t1^3 - 4t1 -1 ). Hmm, not helpful.Alternatively, maybe it's related to the numerator of the first fraction. Let me see:We have ( t1^4 + 4t1^3 - 4t1 -1 ). Let me try to factor it. Maybe try rational roots. Possible rational roots are ±1.Testing t1=1: 1 + 4 -4 -1=0. So, t1=1 is a root. Therefore, (t1 -1) is a factor.Let me perform polynomial division or use synthetic division.Divide ( t1^4 + 4t1^3 - 4t1 -1 ) by (t1 -1):Using synthetic division:Coefficients: 1 (t1^4), 4 (t1^3), 0 (t1^2), -4 (t1), -1 (constant)Write coefficients: 1 | 4 | 0 | -4 | -1Bring down 1.Multiply by 1: 1Add to next coefficient: 4 +1=5Multiply by1:5Add to next coefficient:0 +5=5Multiply by1:5Add to next coefficient: -4 +5=1Multiply by1:1Add to last coefficient: -1 +1=0So, the polynomial factors as (t1 -1)(t1^3 +5t1^2 +5t1 +1)Now, let's factor ( t1^3 +5t1^2 +5t1 +1 ). Try t1=-1:(-1)^3 +5(-1)^2 +5(-1) +1 = -1 +5 -5 +1=0. So, t1=-1 is a root.Thus, factor as (t1 +1)(t1^2 +4t1 +1)Therefore, the denominator factors as:( (t1 -1)(t1 +1)(t1^2 +4t1 +1) )So, denominator is ( (t1 -1)(t1 +1)(t1^2 +4t1 +1) )Now, going back to the slope expression:( m_{NQ} = frac{ -2(t1^2 + 4t1 +1) }{ t1(1 + 2t1) } times frac{ t1^2(1 + 2t1)^2 }{ (t1 -1)(t1 +1)(t1^2 +4t1 +1) } )Notice that ( t1^2 +4t1 +1 ) cancels out in numerator and denominator.So, simplifying:( m_{NQ} = frac{ -2 }{ t1(1 + 2t1) } times frac{ t1^2(1 + 2t1)^2 }{ (t1 -1)(t1 +1) } )Simplify further:- ( t1^2 / t1 = t1 )- ( (1 + 2t1)^2 / (1 + 2t1) = (1 + 2t1) )So,( m_{NQ} = -2 times frac{ t1(1 + 2t1) }{ (t1 -1)(t1 +1) } )Simplify:( m_{NQ} = -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } )So, the slope is ( m_{NQ} = -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } )Now, having the slope, let's write the equation of line NQ. Using point N:( y - yN = m_{NQ}(x - xN) )So,( y - frac{2}{t1} = -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } left( x - frac{1}{t1^2} right) )Let me simplify this equation.First, let me write it as:( y = -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } left( x - frac{1}{t1^2} right) + frac{2}{t1} )Let me distribute the slope:( y = -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } x + 2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } times frac{1}{t1^2} + frac{2}{t1} )Simplify term by term:First term: ( -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } x )Second term: ( 2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } times frac{1}{t1^2} = 2 times frac{ (1 + 2t1) }{ t1(t1^2 -1) } )Third term: ( frac{2}{t1} )So, combining the second and third terms:( frac{2(1 + 2t1)}{ t1(t1^2 -1) } + frac{2}{t1} = frac{2(1 + 2t1) + 2(t1^2 -1)}{ t1(t1^2 -1) } )Simplify numerator:( 2(1 + 2t1) + 2(t1^2 -1) = 2 + 4t1 + 2t1^2 - 2 = 4t1 + 2t1^2 )So,( frac{4t1 + 2t1^2}{ t1(t1^2 -1) } = frac{2t1(2 + t1)}{ t1(t1^2 -1) } = frac{2(2 + t1)}{ t1^2 -1 } )Therefore, the equation of line NQ is:( y = -2 times frac{ t1(1 + 2t1) }{ t1^2 -1 } x + frac{2(2 + t1)}{ t1^2 -1 } )Let me factor out ( frac{2}{ t1^2 -1 } ):( y = frac{2}{ t1^2 -1 } left( - t1(1 + 2t1) x + (2 + t1) right ) )Let me write this as:( y = frac{2}{ t1^2 -1 } left( - t1(1 + 2t1) x + (2 + t1) right ) )Now, to find if this line passes through a fixed point, we need to see if there's a point (a, b) that satisfies the equation for all t1. That is, for some a and b, the equation holds regardless of t1.So, let's set up the equation:( b = frac{2}{ t1^2 -1 } left( - t1(1 + 2t1) a + (2 + t1) right ) )Multiply both sides by ( t1^2 -1 ):( b(t1^2 -1) = 2( - t1(1 + 2t1)a + (2 + t1) ) )Expand both sides:Left side: ( b t1^2 - b )Right side: ( -2a t1(1 + 2t1) + 2(2 + t1) = -2a t1 -4a t1^2 + 4 + 2t1 )So, equation becomes:( b t1^2 - b = -4a t1^2 + (-2a + 2) t1 + 4 )Now, let's collect like terms:Bring all terms to the left side:( b t1^2 - b + 4a t1^2 + (2a - 2) t1 - 4 = 0 )Factor terms by powers of t1:- ( t1^2 ): ( (b + 4a) t1^2 )- ( t1 ): ( (2a - 2) t1 )- Constants: ( -b -4 )So, the equation is:( (b + 4a) t1^2 + (2a - 2) t1 + (-b -4) = 0 )For this equation to hold for all t1, the coefficients of each power of t1 must be zero. Therefore:1. Coefficient of ( t1^2 ): ( b + 4a = 0 )2. Coefficient of ( t1 ): ( 2a - 2 = 0 )3. Constant term: ( -b -4 = 0 )Let's solve these equations:From equation 2: ( 2a - 2 = 0 ) => ( 2a = 2 ) => ( a = 1 )From equation 3: ( -b -4 = 0 ) => ( -b = 4 ) => ( b = -4 )From equation 1: ( b + 4a = 0 ) => ( -4 + 4(1) = 0 ) => ( 0 = 0 ), which holds.So, the fixed point is (a, b) = (1, -4).Therefore, regardless of the value of t1, the line NQ passes through the fixed point (1, -4).That's pretty cool! So, even though Q varies depending on M, the line NQ always goes through (1, -4). I think that's the fixed point we were supposed to find.Let me just recap to make sure I didn't make any mistakes. I parametrized the points on the parabola, found the relationship between t1 and t3 using the condition that MQ passes through B(1, -1), then derived the equation of NQ in terms of t1, and showed that it must pass through (1, -4) for all t1. It seems consistent.I think that's solid. So, the fixed point is (1, -4).**Final Answer**(1) The standard equation of the parabola ( C ) is (boxed{y^2 = 4x}).(2) The line ( NQ ) passes through the fixed point (boxed{(1, -4)}).

answer：First, I need to determine how many ounces of Solution A and Solution B are required to create a 140-ounce mixture that is 80% salt.Let ( x ) represent the ounces of Solution A, which is 40% salt, and ( y ) represent the ounces of Solution B, which is 90% salt.The total volume of the mixture is:[x + y = 140]The total amount of salt in the mixture should be 80% of 140 ounces:[0.40x + 0.90y = 0.80 times 140]Solving the first equation for ( y ):[y = 140 - x]Substituting ( y ) into the second equation:[0.40x + 0.90(140 - x) = 112]Expanding and simplifying:[0.40x + 126 - 0.90x = 112][-0.50x = -14][x = frac{-14}{-0.50} = 28]Therefore, the scientist should use 28 ounces of Solution A.

answer：Okay, so I have this problem here about a line graph. The equation of the line is given as y = mx + b. I remember that in this equation, m is the slope and b is the y-intercept. The problem states that the slope m is positive and specifically it's 3/4. The y-intercept b is given as 2. The question is asking about the product of m and b, which is mb. I need to figure out which of the options A to D is correct regarding this product. The options are:A) mb < 0B) 0 < mb < 1C) mb = 1D) mb > 1Alright, so first, let's recall what the slope and y-intercept represent. The slope, m, tells us how steep the line is and whether it's increasing or decreasing. Since m is positive (3/4), the line slopes upwards from left to right. The y-intercept, b, is where the line crosses the y-axis, which in this case is at 2.Now, the product mb is simply m multiplied by b. So, let's calculate that. Given:m = 3/4b = 2So, mb = (3/4) * 2Let me compute that. Multiplying 3/4 by 2 is the same as 3/4 * 2/1. Multiplying the numerators gives 3*2 = 6, and the denominators give 4*1 = 4. So, 6/4 simplifies to 3/2, which is 1.5.So, mb = 1.5Now, looking at the options:A) mb < 0: Well, 1.5 is definitely not less than 0, so A is incorrect.B) 0 < mb < 1: 1.5 is greater than 1, so it doesn't fall into this range. B is incorrect.C) mb = 1: 1.5 is not equal to 1, so C is incorrect.D) mb > 1: 1.5 is indeed greater than 1, so D is correct.Wait, let me double-check my calculation to be sure I didn't make a mistake. (3/4) * 2 = ?Yes, 3 divided by 4 is 0.75, and 0.75 multiplied by 2 is 1.5. Yep, that's correct.So, the product mb is 1.5, which is greater than 1. Therefore, the correct answer is D.