💡First, I need to determine the perimeter of a semi-circle with a radius of 6.3 cm. The perimeter of a semi-circle consists of two parts: the curved arc and the diameter.To find the length of the curved arc, I'll use the formula for the circumference of a full circle, which is ( C = 2pi r ). Since we only need half of this circumference for the semi-circle, I'll calculate ( pi r ).Next, I'll calculate the diameter of the semi-circle, which is simply twice the radius.Finally, I'll add the length of the curved arc and the diameter to find the total perimeter of the semi-circle.

💡First, I will analyze each number in the set to determine if it is negative.1. **-8**: This is clearly a negative number.2. **0**: Zero is neither positive nor negative.3. **-3²**: I need to evaluate the exponent first. 3 squared is 9, and the negative sign makes it -9, which is negative.4. **-(-5.7)**: This is the negative of a negative number, which results in a positive number, 5.7.After evaluating each number, I find that there are two negative numbers in the set: -8 and -3².

💡To convert the decimal number (500_{10}) to base six, I'll start by identifying the highest power of 6 that fits into 500. Knowing that (6^3 = 216) and (6^4 = 1296), I see that (6^4) is too large, so the highest power needed is (6^3).Next, I'll determine how many times (6^3) fits into 500. It fits 2 times, which gives me the first digit of the base six number. I'll subtract (2 times 216) from 500 to find the remainder, which is 68.Moving to the next lower power, (6^2 = 36), I'll see how many times it fits into the remainder of 68. It fits once, giving me the second digit. Subtracting (1 times 36) from 68 leaves a remainder of 32.Then, I'll consider (6^1 = 6). This fits into 32 five times, providing the third digit. Subtracting (5 times 6) from 32 results in a remainder of 2.Finally, the remaining 2 is less than 6, so it becomes the units digit in the base six number.Putting it all together, the base six representation of (500_{10}) is (2152_6).

💡Alright, so I have this problem here where I'm given a sequence {a_n} with the first term a₁ = 5/2, and a recursive formula a_{n+1}(2 - a_n) = 2 for any natural number n. I need to find the sum of the first 22 terms, S₂₂.Hmm, okay. Let me start by understanding what this recursive formula means. It says that for each term a_{n+1}, it's equal to 2 divided by (2 - a_n). So, basically, each term is determined by the previous term in this way.Let me write that down:a_{n+1} = 2 / (2 - a_n)Given that a₁ = 5/2, I can try to compute the next few terms to see if there's a pattern or if the sequence becomes periodic or something.Starting with a₁ = 5/2.Now, let's compute a₂:a₂ = 2 / (2 - a₁) = 2 / (2 - 5/2) = 2 / (-1/2) = -4Okay, a₂ is -4.Now, a₃:a₃ = 2 / (2 - a₂) = 2 / (2 - (-4)) = 2 / 6 = 1/3Alright, a₃ is 1/3.Next, a₄:a₄ = 2 / (2 - a₃) = 2 / (2 - 1/3) = 2 / (5/3) = 6/5So, a₄ is 6/5.Now, a₅:a₅ = 2 / (2 - a₄) = 2 / (2 - 6/5) = 2 / (4/5) = 10/4 = 5/2Wait a minute, a₅ is 5/2, which is the same as a₁. So, does this mean the sequence repeats every 4 terms? Let me check a₆ to be sure.a₆ = 2 / (2 - a₅) = 2 / (2 - 5/2) = 2 / (-1/2) = -4Yep, a₆ is -4, which is the same as a₂. So, it seems like the sequence is periodic with a period of 4. That is, it repeats every 4 terms: 5/2, -4, 1/3, 6/5, then back to 5/2, and so on.So, knowing that, I can figure out the sum of the first 22 terms by leveraging the periodicity.First, let's find out how many complete cycles of 4 terms are there in 22 terms, and how many extra terms are left.22 divided by 4 is 5 with a remainder of 2. So, there are 5 complete cycles of 4 terms each, and then 2 additional terms.So, S₂₂ = 5*(a₁ + a₂ + a₃ + a₄) + a₁ + a₂Let me compute the sum of one cycle first:Sum of one cycle = a₁ + a₂ + a₃ + a₄ = 5/2 + (-4) + 1/3 + 6/5Hmm, let's compute that step by step.First, convert all fractions to have a common denominator to make addition easier. The denominators are 2, 1, 3, and 5. The least common multiple of 2, 3, and 5 is 30.So, converting each term:5/2 = (5*15)/30 = 75/30-4 = -4/1 = (-4*30)/30 = -120/301/3 = (1*10)/30 = 10/306/5 = (6*6)/30 = 36/30Now, adding them up:75/30 - 120/30 + 10/30 + 36/30Let's compute numerator:75 - 120 + 10 + 36 = (75 + 10 + 36) - 120 = 121 - 120 = 1So, the sum of one cycle is 1/30.Wait, that seems a bit small. Let me double-check my calculations.Wait, 5/2 is 2.5, -4 is -4, 1/3 is approximately 0.333, and 6/5 is 1.2.Adding them up: 2.5 - 4 + 0.333 + 1.2 ≈ (2.5 + 0.333 + 1.2) - 4 ≈ 4.033 - 4 ≈ 0.033.Hmm, 0.033 is approximately 1/30, so that seems correct.So, the sum of one cycle is 1/30.Therefore, 5 cycles would sum up to 5*(1/30) = 5/30 = 1/6.Now, we have 2 extra terms: a₁ and a₂.a₁ is 5/2, a₂ is -4.So, sum of extra terms: 5/2 + (-4) = 5/2 - 4 = (5/2) - (8/2) = (-3/2)Therefore, total sum S₂₂ = 1/6 + (-3/2) = 1/6 - 3/2Convert to common denominator:1/6 - 9/6 = (-8)/6 = (-4)/3So, S₂₂ = -4/3.Wait, that seems a bit negative. Let me check the calculations again.First, the sum of one cycle:5/2 + (-4) + 1/3 + 6/5Compute each term in decimal:5/2 = 2.5-4 = -41/3 ≈ 0.3336/5 = 1.2Adding up: 2.5 - 4 + 0.333 + 1.2 ≈ 2.5 + 0.333 + 1.2 - 4 ≈ 4.033 - 4 ≈ 0.033Which is approximately 1/30, so that's correct.5 cycles: 5*(1/30) = 1/6 ≈ 0.1667Extra terms: a₁ + a₂ = 5/2 + (-4) = 2.5 - 4 = -1.5So, total sum: 0.1667 - 1.5 ≈ -1.3333Which is -4/3, since 4/3 ≈ 1.3333.So, that seems consistent.Therefore, S₂₂ = -4/3.I think that's the answer.

💡Alright, so I have this problem about a parabola in the coordinate plane. It's divided into two parts, and each part has its own subparts. Let me try to understand what's being asked here.First, the setup: Points P and Q are on the parabola C: x² = y, which is a standard upward-opening parabola. Both P and Q are distinct from the origin O. There's a line l that passes through point P and is tangent to the parabola C. This line l intersects the x-axis at point M. Additionally, the product of the slopes of lines OQ and l is -2.Part (1) asks me to prove that the line PQ passes through a fixed point D and to find the coordinates of D.Okay, so for part (1), I need to show that no matter where P and Q are on the parabola (satisfying the given conditions), the line connecting P and Q will always pass through a specific point D. Then, I have to find where D is located.Let me start by recalling some properties of parabolas. The equation is x² = y, so it's symmetric about the y-axis. The derivative, which gives the slope of the tangent at any point, is dy/dx = 2x. So, if P is a point (a, a²) on the parabola, the slope of the tangent at P is 2a.Given that line l is tangent at P, its slope is 2a. The product of the slopes of OQ and l is -2. So, if the slope of OQ is m, then m * (2a) = -2, which implies m = -1/a.But OQ is the line from the origin O(0,0) to Q, which is another point on the parabola. Let's denote Q as (b, b²). The slope of OQ is (b² - 0)/(b - 0) = b. So, the slope of OQ is b. Therefore, from the product condition, b * (2a) = -2, which simplifies to ab = -1. So, b = -1/a.So, if P is (a, a²), then Q is (-1/a, ( -1/a )²) = (-1/a, 1/a²).Now, I need to find the equation of the line PQ. Let's compute the slope of PQ first.The slope of PQ is (y_Q - y_P)/(x_Q - x_P) = (1/a² - a²)/(-1/a - a).Let me compute numerator and denominator separately:Numerator: 1/a² - a² = (1 - a⁴)/a²Denominator: -1/a - a = -(1 + a²)/aSo, the slope is [(1 - a⁴)/a²] / [-(1 + a²)/a] = [(1 - a⁴)/a²] * [ -a/(1 + a²) ] = - (1 - a⁴) / [a(1 + a²)]Simplify numerator: 1 - a⁴ = (1 - a²)(1 + a²)So, slope = - (1 - a²)(1 + a²) / [a(1 + a²)] = - (1 - a²)/a = (a² - 1)/aSo, the slope of PQ is (a² - 1)/a.Now, let's write the equation of PQ. Using point P(a, a²):y - a² = [(a² - 1)/a](x - a)Let me simplify this:y = [(a² - 1)/a]x - [(a² - 1)/a]*a + a²Simplify the second term: - (a² - 1) + a² = -a² + 1 + a² = 1So, the equation of PQ is y = [(a² - 1)/a]x + 1Interesting. So, regardless of the value of a, the equation of PQ is y = [(a² - 1)/a]x + 1. This suggests that when x = 0, y = 1. Therefore, the line PQ always passes through the point (0,1).Therefore, the fixed point D is (0,1).Wait, let me double-check this. If I plug x=0 into the equation, y = 1, regardless of a. So, yes, the line PQ always passes through (0,1). So, D is (0,1).That seems straightforward. So, part (1) is done.Moving on to part (2). It's more involved.Part (2): Let A and B be the two points where the line passing through M and perpendicular to l intersects the ellipse x²/4 + y² = 1. Let H be the point where the line passing through D and parallel to l intersects the line AB. Denote the area of triangle OPQ as S and the area of triangle ABD as T.(i) When T/S² takes the maximum value, find the ordinate of point P.(ii) Prove that there exists a fixed point G such that |GH| is a constant value.Alright, so part (2) has two subparts. Let me start with part (i).First, let's parse the problem.We have point M, which is where the tangent line l at P intersects the x-axis. Then, we draw a line through M that's perpendicular to l. This line intersects the ellipse x²/4 + y² = 1 at points A and B.Then, we have point H, which is the intersection of two lines: one is the line through D (which we found as (0,1)) parallel to l, and the other is line AB.We need to find the area S of triangle OPQ and the area T of triangle ABD, then find when T/S² is maximized, and find the y-coordinate (ordinate) of point P.Let me start by finding expressions for all these points and areas.First, let's recap:Point P is (a, a²). The tangent at P has slope 2a, so its equation is y = 2a(x - a) + a², which simplifies to y = 2a x - 2a² + a² = 2a x - a².This line intersects the x-axis at M, where y=0:0 = 2a x - a² => x = a/2. So, M is (a/2, 0).Now, the line perpendicular to l through M: since l has slope 2a, the perpendicular has slope -1/(2a). So, the equation is y - 0 = (-1/(2a))(x - a/2), which simplifies to y = (-1/(2a))x + 1/(4).This line intersects the ellipse x²/4 + y² = 1.Let me write the equation of the ellipse: x²/4 + y² = 1.Substitute y from the perpendicular line into the ellipse equation:x²/4 + [ (-1/(2a))x + 1/4 ]² = 1Let me compute [ (-1/(2a))x + 1/4 ]²:= (1/(4a²))x² - (1/(4a))(x) + 1/16So, plugging into the ellipse equation:x²/4 + (1/(4a²))x² - (1/(4a))x + 1/16 = 1Multiply through by 16a² to eliminate denominators:4a² x² + 4x² - 4a x + a² = 16a²Combine like terms:(4a² + 4)x² - 4a x + (a² - 16a²) = 0Simplify:4(a² + 1)x² - 4a x - 15a² = 0Divide through by 4:(a² + 1)x² - a x - (15/4)a² = 0So, quadratic in x: (a² + 1)x² - a x - (15/4)a² = 0Let me denote this as:A x² + B x + C = 0, where A = a² + 1, B = -a, C = -15a²/4The solutions are x = [a ± sqrt(a² + 4*(a² + 1)*(15a²/4))]/(2*(a² + 1))Wait, let me compute discriminant D:D = B² - 4AC = a² - 4*(a² + 1)*(-15a²/4)Simplify:= a² + 15a²(a² + 1)= a² + 15a⁴ + 15a²= 15a⁴ + 16a²So, sqrt(D) = sqrt(15a⁴ + 16a²) = a*sqrt(15a² + 16)Thus, x = [a ± a*sqrt(15a² + 16)]/(2*(a² + 1)) = a[1 ± sqrt(15a² + 16)]/(2*(a² + 1))Therefore, the x-coordinates of A and B are:x₁ = a[1 + sqrt(15a² + 16)]/(2*(a² + 1))x₂ = a[1 - sqrt(15a² + 16)]/(2*(a² + 1))Corresponding y-coordinates can be found from the line equation y = (-1/(2a))x + 1/4.So, y₁ = (-1/(2a))x₁ + 1/4Similarly, y₂ = (-1/(2a))x₂ + 1/4But perhaps I don't need the exact coordinates of A and B, but rather the length AB or something related to triangle ABD.Wait, triangle ABD: points A, B, D. D is (0,1). So, to find the area T, I need coordinates of A and B, then compute the area.Alternatively, maybe I can find the length AB and the height from D to AB, but since AB is a chord of the ellipse, perhaps there's a smarter way.But let me think step by step.First, I need to find the area S of triangle OPQ.Points O(0,0), P(a, a²), Q(-1/a, 1/a²).The area can be found using determinant formula:S = (1/2)| (x_P y_Q - x_Q y_P) | = (1/2)| a*(1/a²) - (-1/a)*(a²) | = (1/2)| 1/a + a | = (1/2)|a + 1/a|Since a ≠ 0, and points are distinct from O, a ≠ ±1? Wait, no, a can be any real except 0.But S is positive, so S = (1/2)(|a + 1/a|). Since a + 1/a can be positive or negative, but squared, so S² = (1/4)(a + 1/a)².Wait, actually, since S is an area, it's always positive, so S = (1/2)|a + 1/a|, and S² = (1/4)(a + 1/a)².Now, moving on to T, the area of triangle ABD.Points A, B, D(0,1). To find the area, I can use the determinant formula as well.But to do that, I need coordinates of A and B. Alternatively, perhaps I can find the length of AB and the height from D to AB.But maybe it's easier to parametrize.Alternatively, since AB is a chord of the ellipse, and D is fixed, perhaps the area can be expressed in terms of the coordinates.But let me try to find coordinates of A and B.From earlier, we have x-coordinates of A and B:x₁ = a[1 + sqrt(15a² + 16)]/(2*(a² + 1))x₂ = a[1 - sqrt(15a² + 16)]/(2*(a² + 1))Similarly, y₁ = (-1/(2a))x₁ + 1/4Similarly, y₂ = (-1/(2a))x₂ + 1/4So, let me compute y₁ and y₂.Compute y₁:= (-1/(2a)) * [a(1 + sqrt(15a² + 16))/(2(a² + 1))] + 1/4Simplify:= [ - (1 + sqrt(15a² + 16)) / (4(a² + 1)) ] + 1/4Similarly, y₂:= (-1/(2a)) * [a(1 - sqrt(15a² + 16))/(2(a² + 1))] + 1/4= [ - (1 - sqrt(15a² + 16)) / (4(a² + 1)) ] + 1/4So, points A and B are:A: ( x₁, y₁ ) = ( a[1 + sqrt(15a² + 16)]/(2(a² + 1)), [ - (1 + sqrt(15a² + 16)) / (4(a² + 1)) + 1/4 ] )B: ( x₂, y₂ ) = ( a[1 - sqrt(15a² + 16)]/(2(a² + 1)), [ - (1 - sqrt(15a² + 16)) / (4(a² + 1)) + 1/4 ] )This looks complicated. Maybe I can simplify y₁ and y₂.Let me compute y₁:= [ - (1 + sqrt(15a² + 16)) / (4(a² + 1)) ] + 1/4= [ -1 - sqrt(15a² + 16) + (a² + 1) ] / [4(a² + 1)]= [ (a² + 1) - 1 - sqrt(15a² + 16) ] / [4(a² + 1)]= [ a² - sqrt(15a² + 16) ] / [4(a² + 1)]Similarly, y₂:= [ - (1 - sqrt(15a² + 16)) / (4(a² + 1)) ] + 1/4= [ -1 + sqrt(15a² + 16) + (a² + 1) ] / [4(a² + 1)]= [ a² + sqrt(15a² + 16) ] / [4(a² + 1)]So, points A and B are:A: ( a[1 + sqrt(15a² + 16)]/(2(a² + 1)), [ a² - sqrt(15a² + 16) ] / [4(a² + 1)] )B: ( a[1 - sqrt(15a² + 16)]/(2(a² + 1)), [ a² + sqrt(15a² + 16) ] / [4(a² + 1)] )Now, to find the area T of triangle ABD, where D is (0,1).Using the determinant formula for area:T = (1/2)| x_A(y_B - y_D) + x_B(y_D - y_A) + x_D(y_A - y_B) |Since x_D = 0, the last term is 0.So, T = (1/2)| x_A(y_B - 1) + x_B(1 - y_A) |Let me compute this.First, compute y_B - 1:= [ a² + sqrt(15a² + 16) ] / [4(a² + 1)] - 1= [ a² + sqrt(15a² + 16) - 4(a² + 1) ] / [4(a² + 1)]= [ -3a² + sqrt(15a² + 16) - 4 ] / [4(a² + 1)]Similarly, 1 - y_A:= 1 - [ a² - sqrt(15a² + 16) ] / [4(a² + 1) ]= [4(a² + 1) - a² + sqrt(15a² + 16) ] / [4(a² + 1)]= [3a² + 4 + sqrt(15a² + 16) ] / [4(a² + 1)]Now, x_A = a[1 + sqrt(15a² + 16)]/(2(a² + 1))x_B = a[1 - sqrt(15a² + 16)]/(2(a² + 1))So, plug into T:T = (1/2)| x_A(y_B - 1) + x_B(1 - y_A) |= (1/2)| [a(1 + s)/(2(a² + 1))] * [ (-3a² + s - 4)/(4(a² + 1)) ] + [a(1 - s)/(2(a² + 1))] * [ (3a² + 4 + s)/(4(a² + 1)) ] |, where s = sqrt(15a² + 16)This is getting really messy. Maybe there's a better approach.Alternatively, perhaps I can parametrize the line AB and find its equation, then find H, then compute areas.Wait, but H is the intersection of line AB and the line through D parallel to l.Line l has slope 2a, so the line through D parallel to l has slope 2a. So, its equation is y - 1 = 2a(x - 0), which is y = 2a x + 1.So, line AB is y = (-1/(2a))x + 1/4, as we had earlier.Wait, no, line AB is the same as the line through M perpendicular to l, which is y = (-1/(2a))x + 1/4.So, line AB: y = (-1/(2a))x + 1/4Line through D parallel to l: y = 2a x + 1Intersection point H is where these two lines meet.So, set 2a x + 1 = (-1/(2a))x + 1/4Multiply both sides by 2a to eliminate denominators:4a² x + 2a = -x + (1/4)*2aSimplify:4a² x + 2a = -x + (a/2)Bring all terms to left:4a² x + x + 2a - a/2 = 0Factor x:x(4a² + 1) + (3a/2) = 0Thus, x = - (3a/2) / (4a² + 1)Then, y = 2a x + 1 = 2a*(-3a/(2(4a² + 1))) + 1 = -3a²/(4a² + 1) + 1 = ( -3a² + 4a² + 1 ) / (4a² + 1 ) = (a² + 1)/(4a² + 1)So, point H is ( -3a/(2(4a² + 1)), (a² + 1)/(4a² + 1) )Now, to find area T of triangle ABD.Points A, B, D(0,1). Let me use the shoelace formula.But since I have coordinates of A and B, and D, perhaps it's manageable.Alternatively, since AB is a line segment, and D is a point, the area can be found as (1/2)*base*height, where base is AB and height is the distance from D to line AB.But line AB is y = (-1/(2a))x + 1/4.The distance from D(0,1) to this line is | (-1/(2a))*0 - 1 + 1/4 | / sqrt( ( -1/(2a) )² + 1 )= | -1 + 1/4 | / sqrt( 1/(4a²) + 1 )= | -3/4 | / sqrt( (1 + 4a²)/(4a²) )= (3/4) / ( sqrt(1 + 4a²)/(2|a|) )= (3/4) * (2|a|)/sqrt(1 + 4a²)= (3|a|)/(2 sqrt(1 + 4a²))But since a is a real number (except 0), and we're dealing with areas, we can consider a > 0 without loss of generality, because the problem is symmetric in a and -a.So, distance = (3a)/(2 sqrt(1 + 4a²))Now, the length of AB can be found from the ellipse equation.But since AB is a chord of the ellipse x²/4 + y² = 1, and we have the line y = (-1/(2a))x + 1/4 intersecting it.Alternatively, we can compute the length AB from the coordinates of A and B.But given that we have the quadratic equation earlier, we can use the formula for the length of a chord in terms of the quadratic solutions.Recall that for a quadratic equation Ax² + Bx + C = 0, the difference of roots is sqrt(D)/A, where D is discriminant.But actually, the distance between A and B can be found using the distance formula:AB = sqrt( (x₁ - x₂)² + (y₁ - y₂)² )From earlier, x₁ - x₂ = [a(1 + s) - a(1 - s)] / [2(a² + 1)] = [2a s] / [2(a² + 1)] = a s / (a² + 1), where s = sqrt(15a² + 16)Similarly, y₁ - y₂ = [ (a² - s ) - (a² + s) ] / [4(a² + 1)] = (-2s)/[4(a² + 1)] = -s/(2(a² + 1))So, AB = sqrt( [a s / (a² + 1)]² + [ -s/(2(a² + 1)) ]² )= sqrt( a² s² / (a² + 1)² + s² / (4(a² + 1)² ) )= sqrt( [4a² s² + s² ] / [4(a² + 1)² ] )= sqrt( s² (4a² + 1) / [4(a² + 1)² ] )= (s sqrt(4a² + 1)) / [2(a² + 1)]But s = sqrt(15a² + 16), so:AB = sqrt(15a² + 16) * sqrt(4a² + 1) / [2(a² + 1)]Therefore, AB = sqrt( (15a² + 16)(4a² + 1) ) / [2(a² + 1)]So, the area T is (1/2)*AB*distance from D to AB.We have AB as above, and distance as (3a)/(2 sqrt(1 + 4a²))Thus, T = (1/2) * [ sqrt( (15a² + 16)(4a² + 1) ) / (2(a² + 1)) ] * [ 3a / (2 sqrt(1 + 4a²)) ]Simplify:= (1/2) * [ sqrt( (15a² + 16)(4a² + 1) ) / (2(a² + 1)) ] * [ 3a / (2 sqrt(4a² + 1)) ]= (1/2) * [ sqrt(15a² + 16) * sqrt(4a² + 1) / (2(a² + 1)) ] * [ 3a / (2 sqrt(4a² + 1)) ]Notice that sqrt(4a² + 1) cancels out:= (1/2) * [ sqrt(15a² + 16) / (2(a² + 1)) ] * [ 3a / 2 ]= (1/2) * [ 3a sqrt(15a² + 16) ] / [4(a² + 1) ]= (3a sqrt(15a² + 16)) / [8(a² + 1) ]So, T = (3a sqrt(15a² + 16)) / [8(a² + 1) ]Earlier, S = (1/2)|a + 1/a|. Since we're considering a > 0, S = (1/2)(a + 1/a). Thus, S² = (1/4)(a + 1/a)² = (1/4)(a² + 2 + 1/a²)Therefore, T/S² = [ (3a sqrt(15a² + 16)) / (8(a² + 1)) ] / [ (1/4)(a² + 2 + 1/a²) ]Simplify:= [ (3a sqrt(15a² + 16)) / (8(a² + 1)) ] * [ 4 / (a² + 2 + 1/a²) ]= [ 3a sqrt(15a² + 16) * 4 ] / [8(a² + 1)(a² + 2 + 1/a²) ]Simplify constants:= [ 12a sqrt(15a² + 16) ] / [8(a² + 1)(a² + 2 + 1/a²) ]= [ 3a sqrt(15a² + 16) ] / [2(a² + 1)(a² + 2 + 1/a²) ]Now, let me simplify the denominator:a² + 2 + 1/a² = (a^4 + 2a² + 1)/a² = (a² + 1)^2 / a²So, denominator becomes:2(a² + 1) * (a² + 1)^2 / a² = 2(a² + 1)^3 / a²Thus, T/S² = [ 3a sqrt(15a² + 16) ] / [ 2(a² + 1)^3 / a² ]= [ 3a sqrt(15a² + 16) * a² ] / [ 2(a² + 1)^3 ]= [ 3a³ sqrt(15a² + 16) ] / [ 2(a² + 1)^3 ]Let me denote t = a², so t > 0.Then, T/S² = [ 3 t^(3/2) sqrt(15t + 16) ] / [ 2(t + 1)^3 ]= [ 3 sqrt(t³(15t + 16)) ] / [ 2(t + 1)^3 ]= [ 3 sqrt(15t⁴ + 16t³) ] / [ 2(t + 1)^3 ]So, T/S² = (3/2) * sqrt(15t⁴ + 16t³) / (t + 1)^3Let me write this as:T/S² = (3/2) * sqrt(t³(15t + 16)) / (t + 1)^3= (3/2) * t^(3/2) sqrt(15t + 16) / (t + 1)^3To maximize T/S², we can consider the function f(t) = sqrt(15t⁴ + 16t³) / (t + 1)^3, and find its maximum.Alternatively, square both sides to make it easier:Let me define f(t) = (T/S²)^2 = (9/4) * (15t⁴ + 16t³) / (t + 1)^6So, f(t) = (9/4) * (15t⁴ + 16t³) / (t + 1)^6To maximize f(t), we can take derivative and set to zero.Let me compute f(t):f(t) = (9/4) * t³(15t + 16) / (t + 1)^6Let me set u = t, so f(u) = (9/4) * u³(15u + 16) / (u + 1)^6To find maximum, compute f’(u) and set to zero.Let me compute derivative using quotient rule.Let me denote numerator N = u³(15u + 16) = 15u⁴ + 16u³Denominator D = (u + 1)^6Then, f(u) = (9/4) * N / Df’(u) = (9/4) * (N’ D - N D’) / D²Compute N’:N’ = 60u³ + 48u²Compute D’:D’ = 6(u + 1)^5Thus,f’(u) = (9/4) * [ (60u³ + 48u²)(u + 1)^6 - (15u⁴ + 16u³)*6(u + 1)^5 ] / (u + 1)^12Factor out common terms:= (9/4) * (u + 1)^5 [ (60u³ + 48u²)(u + 1) - 6(15u⁴ + 16u³) ] / (u + 1)^12Simplify denominator:= (9/4) * [ (60u³ + 48u²)(u + 1) - 6(15u⁴ + 16u³) ] / (u + 1)^7Now, expand numerator:First term: (60u³ + 48u²)(u + 1) = 60u⁴ + 60u³ + 48u³ + 48u² = 60u⁴ + 108u³ + 48u²Second term: -6(15u⁴ + 16u³) = -90u⁴ - 96u³Combine:60u⁴ + 108u³ + 48u² -90u⁴ -96u³ = (-30u⁴) + 12u³ + 48u²Factor:= -30u⁴ + 12u³ + 48u² = -6u²(5u² - 2u - 8)Wait, let me factor:-30u⁴ + 12u³ + 48u² = -6u²(5u² - 2u - 8)Wait, 5u² - 2u -8 factors as (5u + 8)(u - 1), but let me check:5u² - 2u -8: discriminant = 4 + 160 = 164, which is not a perfect square, so it doesn't factor nicely.Alternatively, perhaps I made a miscalculation.Wait, let me recompute:Numerator after expansion:60u⁴ + 108u³ + 48u² -90u⁴ -96u³= (60u⁴ -90u⁴) + (108u³ -96u³) + 48u²= (-30u⁴) + (12u³) + 48u²Yes, that's correct.So, f’(u) = (9/4) * [ -30u⁴ + 12u³ + 48u² ] / (u + 1)^7Factor numerator:= (9/4) * [ -6u²(5u² - 2u - 8) ] / (u + 1)^7Set f’(u) = 0:-6u²(5u² - 2u - 8) = 0Solutions: u = 0 or 5u² - 2u -8 = 0But u = t = a² > 0, so u ≠ 0.Solve 5u² - 2u -8 = 0:Using quadratic formula:u = [2 ± sqrt(4 + 160)] / 10 = [2 ± sqrt(164)] / 10 = [2 ± 2sqrt(41)] / 10 = [1 ± sqrt(41)] / 5Since u > 0, we take the positive root:u = [1 + sqrt(41)] / 5Thus, t = [1 + sqrt(41)] / 5Therefore, a² = t = [1 + sqrt(41)] / 5So, the ordinate of point P is y_P = a² = [1 + sqrt(41)] / 5Hence, when T/S² is maximized, the ordinate of P is [1 + sqrt(41)] / 5Now, moving to part (2)(ii):Prove that there exists a fixed point G such that |GH| is a constant value.From earlier, point H is the intersection of line AB and the line through D parallel to l.We found H as ( -3a/(2(4a² + 1)), (a² + 1)/(4a² + 1) )We need to show that there exists a fixed point G such that |GH| is constant.Let me think about the locus of H as a varies.Given that H is expressed in terms of a, perhaps we can find a relation that shows H lies on a circle or some conic, and G is the center or something.Alternatively, perhaps H lies on a circle with center G and fixed radius.Let me compute coordinates of H:x_H = -3a/(2(4a² + 1))y_H = (a² + 1)/(4a² + 1)Let me denote s = a², so s > 0.Then,x_H = -3a/(2(4s + 1))But a = sqrt(s) or -sqrt(s), but since we're dealing with coordinates, x_H can be positive or negative depending on a.But perhaps I can express x_H and y_H in terms of s.Wait, but s = a², so a = sqrt(s), but x_H involves a, not s.Alternatively, let me express x_H in terms of s:x_H = -3a/(2(4s + 1)) = -3 sqrt(s)/(2(4s + 1))But this complicates things.Alternatively, let me consider parametric equations.Let me set t = a, so x_H = -3t/(2(4t² + 1)), y_H = (t² + 1)/(4t² + 1)Let me see if I can eliminate t to find a relation between x and y.Let me denote x = x_H, y = y_H.So,x = -3t/(2(4t² + 1)) => x = -3t/(8t² + 2)y = (t² + 1)/(4t² + 1)Let me solve for t from x:From x = -3t/(8t² + 2), multiply both sides by (8t² + 2):x(8t² + 2) = -3t=> 8x t² + 2x = -3t=> 8x t² + 3t + 2x = 0This is quadratic in t:8x t² + 3t + 2x = 0Let me solve for t:t = [ -3 ± sqrt(9 - 64x²) ] / (16x)But this seems complicated.Alternatively, let me express y in terms of t:y = (t² + 1)/(4t² + 1) = [ (t² + 1) ] / [4t² + 1 ]Let me write y = (t² + 1)/(4t² + 1) => y(4t² + 1) = t² + 1 => 4y t² + y = t² + 1 => (4y - 1)t² + (y - 1) = 0So, (4y - 1)t² = 1 - y => t² = (1 - y)/(4y - 1)But from x = -3t/(8t² + 2), let me express t in terms of x and y.Wait, perhaps another approach.Let me consider that H lies on the circle with diameter DN, where N is the midpoint of AB or something. But I'm not sure.Wait, from earlier, line DH is parallel to l, which has slope 2a. So, DH has slope 2a.But line AB has slope -1/(2a), so DH is perpendicular to AB.Wait, if DH is perpendicular to AB, and H lies on AB, then H is the foot of the perpendicular from D to AB.But wait, no, because DH is parallel to l, which is not necessarily perpendicular to AB.Wait, line AB has slope -1/(2a), and DH has slope 2a, which is negative reciprocal, so they are perpendicular.Ah, yes! Because 2a * (-1/(2a)) = -1, so DH is perpendicular to AB.Therefore, H is the foot of the perpendicular from D to AB.But since AB is a chord of the ellipse, and D is fixed, perhaps H lies on a circle with diameter DN, where N is some fixed point.Alternatively, since DH is perpendicular to AB, and AB is variable, perhaps H lies on a circle with center at the midpoint of OD or something.Wait, O is (0,0), D is (0,1). Midpoint is (0, 0.5). Maybe H lies on a circle centered at (0, 0.5) with some radius.Let me compute |GH| where G is (0, 0.5).Compute distance from H to G(0, 0.5):|GH| = sqrt( x_H² + (y_H - 0.5)^2 )From H's coordinates:x_H = -3a/(2(4a² + 1))y_H = (a² + 1)/(4a² + 1)Compute y_H - 0.5:= (a² + 1)/(4a² + 1) - 1/2 = [2(a² + 1) - (4a² + 1)] / [2(4a² + 1)] = [2a² + 2 - 4a² -1] / [2(4a² + 1)] = (-2a² +1)/[2(4a² +1)]So,|GH|² = x_H² + (y_H - 0.5)^2= [9a² / (4(4a² +1)^2)] + [ ( -2a² +1 )² / (4(4a² +1)^2 ) ]= [9a² + (4a⁴ -4a² +1)] / [4(4a² +1)^2 ]= [4a⁴ +5a² +1] / [4(4a² +1)^2 ]Factor numerator:4a⁴ +5a² +1 = (4a² +1)(a² +1)Thus,|GH|² = (4a² +1)(a² +1) / [4(4a² +1)^2 ] = (a² +1) / [4(4a² +1) ]But wait, this is not a constant. Hmm.Wait, but let me compute |GH|:|GH| = sqrt( (a² +1)/(4(4a² +1)) ) = sqrt( (a² +1)/(16a² +4) ) = sqrt( (a² +1)/(4(4a² +1)) ) = (1/2) sqrt( (a² +1)/(4a² +1) )This is not constant. So, perhaps G is not (0, 0.5).Alternatively, maybe G is another point.Wait, let me think differently.Since H is the foot of the perpendicular from D to AB, and AB is a chord of the ellipse, perhaps H lies on the ellipse's director circle or something, but I'm not sure.Alternatively, perhaps H lies on a circle centered at some point G, and |GH| is constant.Let me try to find such a point G.Suppose G is (h, k). Then, |GH|² = (x_H - h)^2 + (y_H - k)^2 is constant.We need to find h and k such that this expression is independent of a.From H's coordinates:x_H = -3a/(2(4a² +1))y_H = (a² +1)/(4a² +1)So,|GH|² = [ -3a/(2(4a² +1)) - h ]² + [ (a² +1)/(4a² +1) - k ]²Let me denote s = a², so a = sqrt(s), but this might complicate things. Alternatively, let me express in terms of t = a.Let me set t = a, so:x_H = -3t/(2(4t² +1))y_H = (t² +1)/(4t² +1)So,|GH|² = [ -3t/(2(4t² +1)) - h ]² + [ (t² +1)/(4t² +1) - k ]²Let me expand this:= [ (-3t - 2h(4t² +1))² / (4(4t² +1)^2) ] + [ (t² +1 - k(4t² +1))² / (4t² +1)^2 ]This is quite complicated. Maybe instead, I can assume that |GH| is constant, say c, and find h and k such that the equation holds for all t.But this seems difficult. Alternatively, perhaps G is the midpoint of OD, which is (0, 0.5). But earlier, |GH| wasn't constant.Alternatively, maybe G is the center of the ellipse, which is (0,0). Let me compute |GH|:|GH|² = x_H² + y_H²= [9a² / (4(4a² +1)^2)] + [ (a² +1)^2 / (4a² +1)^2 ]= [9a² + 4(a² +1)^2 ] / [4(4a² +1)^2 ]Expand numerator:= 9a² + 4(a⁴ + 2a² +1 ) = 9a² +4a⁴ +8a² +4 = 4a⁴ +17a² +4Denominator: 4(4a² +1)^2 = 4(16a⁴ +8a² +1 ) = 64a⁴ +32a² +4So, |GH|² = (4a⁴ +17a² +4)/(64a⁴ +32a² +4)This is not constant.Alternatively, maybe G is (0, something else). Let me try to find h and k such that |GH|² is constant.Let me denote |GH|² = c², so:[ (-3t/(2(4t² +1)) - h )² + ( (t² +1)/(4t² +1) - k )² ] = c²This must hold for all t. Let me expand and collect like terms.This seems too involved. Maybe another approach.Wait, from earlier, we have:x_H = -3a/(2(4a² +1)) = -3a/(8a² +2)y_H = (a² +1)/(4a² +1)Let me denote u = a, so:x = -3u/(8u² +2)y = (u² +1)/(4u² +1)Let me try to eliminate u.From x = -3u/(8u² +2), let me solve for u:x(8u² +2) = -3u => 8x u² + 2x + 3u = 0This is quadratic in u:8x u² + 3u + 2x = 0Let me solve for u:u = [ -3 ± sqrt(9 - 64x²) ] / (16x)But this seems messy.Alternatively, let me express y in terms of x.From y = (u² +1)/(4u² +1)Let me write y = (u² +1)/(4u² +1) => y(4u² +1) = u² +1 => 4y u² + y = u² +1 => (4y -1)u² = 1 - y => u² = (1 - y)/(4y -1)From x = -3u/(8u² +2), let me express u in terms of x and y.But u² = (1 - y)/(4y -1), so u = sqrt( (1 - y)/(4y -1) ) or negative.But x = -3u/(8u² +2) = -3u/(8*(1 - y)/(4y -1) +2 )Simplify denominator:= 8*(1 - y)/(4y -1) +2 = [8(1 - y) + 2(4y -1)] / (4y -1) = [8 -8y +8y -2]/(4y -1) = (6)/(4y -1)Thus, x = -3u / (6/(4y -1)) ) = -3u*(4y -1)/6 = -u*(4y -1)/2So, x = -u*(4y -1)/2But u = sqrt( (1 - y)/(4y -1) ) or negative.Let me take u = sqrt( (1 - y)/(4y -1) )Then,x = -sqrt( (1 - y)/(4y -1) )*(4y -1)/2 = -sqrt( (1 - y)(4y -1) ) / 2But this is getting too complicated. Maybe instead, consider that H lies on a circle.Let me assume that H lies on a circle with center G(h,k) and radius r.Then, (x_H - h)^2 + (y_H - k)^2 = r² for all a.We need to find h, k, r such that this holds.From H's coordinates:x_H = -3a/(2(4a² +1))y_H = (a² +1)/(4a² +1)Let me plug into the circle equation:[ -3a/(2(4a² +1)) - h ]² + [ (a² +1)/(4a² +1) - k ]² = r²Let me expand this:= [ ( -3a - 2h(4a² +1) )² / (4(4a² +1)^2) ] + [ (a² +1 - k(4a² +1) )² / (4a² +1)^2 ] = r²Multiply both sides by (4a² +1)^2:[ ( -3a - 2h(4a² +1) )² /4 ] + [ (a² +1 - k(4a² +1) )² ] = r²(4a² +1)^2This must hold for all a, so coefficients of powers of a must match on both sides.This seems very involved, but let me try to equate coefficients.Let me denote:First term: [ ( -3a - 2h(4a² +1) )² ] /4= [ 9a² + 12h a(4a² +1) + 4h²(4a² +1)^2 ] /4= (9a²)/4 + 3h a(4a² +1) + h²(4a² +1)^2Second term: [ (a² +1 - k(4a² +1) )² ]= (a² +1)^2 - 2k(a² +1)(4a² +1) + k²(4a² +1)^2So, total left side:= (9a²)/4 + 3h a(4a² +1) + h²(4a² +1)^2 + (a² +1)^2 - 2k(a² +1)(4a² +1) + k²(4a² +1)^2Now, expand each term:1. (9a²)/42. 3h a(4a² +1) = 12h a³ + 3h a3. h²(4a² +1)^2 = h²(16a⁴ +8a² +1)4. (a² +1)^2 = a⁴ + 2a² +15. -2k(a² +1)(4a² +1) = -2k(4a⁴ +5a² +1)6. k²(4a² +1)^2 = k²(16a⁴ +8a² +1)Now, combine all terms:= (9a²)/4 + 12h a³ + 3h a + h²(16a⁴ +8a² +1) + a⁴ + 2a² +1 -2k(4a⁴ +5a² +1) + k²(16a⁴ +8a² +1)Now, collect like terms by power of a:a⁴ terms:h²*16a⁴ + a⁴ -2k*4a⁴ +k²*16a⁴ = (16h² +1 -8k +16k²)a⁴a³ terms:12h a³a² terms:(9/4)a² + h²*8a² + 2a² -2k*5a² +k²*8a² = (9/4 +8h² +2 -10k +8k²)a²a terms:3h aconstant terms:h²*1 +1 -2k*1 +k²*1 = (h² +1 -2k +k²)So, left side becomes:(16h² +1 -8k +16k²)a⁴ +12h a³ + (9/4 +8h² +2 -10k +8k²)a² +3h a + (h² +1 -2k +k²)This must equal right side:r²(4a² +1)^2 = r²(16a⁴ +8a² +1)So, equate coefficients:For a⁴:16h² +1 -8k +16k² = 16r²For a³:12h = 0 => h = 0For a²:9/4 +8h² +2 -10k +8k² =8r²For a:3h =0 => h=0 (consistent)For constants:h² +1 -2k +k² = r²From a³ term, h=0.So, h=0.Now, substitute h=0 into other equations.For a⁴:16*0 +1 -8k +16k² =16r² =>1 -8k +16k² =16r²For a²:9/4 +0 +2 -10k +8k² =8r² => (9/4 +2) -10k +8k² =8r² => (17/4) -10k +8k² =8r²For constants:0 +1 -2k +k² = r² =>1 -2k +k² = r²Now, we have three equations:1. 1 -8k +16k² =16r²2. 17/4 -10k +8k² =8r²3. 1 -2k +k² = r²Let me express r² from equation 3: r² =1 -2k +k²Plug into equation 1:1 -8k +16k² =16(1 -2k +k²) =16 -32k +16k²Simplify:1 -8k +16k² =16 -32k +16k²Subtract 16k² from both sides:1 -8k =16 -32kBring all terms to left:1 -8k -16 +32k =0 =>24k -15=0 =>24k=15 =>k=15/24=5/8So, k=5/8Now, from equation 3:r²=1 -2*(5/8) + (5/8)^2=1 -5/4 +25/64= (64/64 -80/64 +25/64)=9/64Thus, r=3/8Now, check equation 2:17/4 -10k +8k²=8r²Plug k=5/8, r²=9/64:Left side:17/4 -10*(5/8) +8*(25/64)=17/4 -50/8 +200/64=17/4 -25/4 +25/8= (17 -25)/4 +25/8= (-8)/4 +25/8= -2 +25/8= (-16/8 +25/8)=9/8Right side:8*(9/64)=72/64=9/8Thus, equation holds.Therefore, the circle equation is:(x -0)^2 + (y -5/8)^2 = (3/8)^2So, H lies on the circle centered at G(0,5/8) with radius 3/8.Therefore, |GH|=3/8, which is constant.Hence, there exists a fixed point G(0,5/8) such that |GH|=3/8, a constant.So, part (2)(ii) is proved.

💡To determine the average number of minutes run per day by the sixth, seventh, and eighth graders, we start by defining the number of students in each grade. Let ( e ) represent the number of eighth graders. According to the problem:- The number of sixth graders is three times the number of eighth graders, which is ( 3e ).- The number of seventh graders is twice the number of eighth graders, which is ( 2e ).Next, we calculate the total minutes run by each grade:- Sixth graders run an average of 20 minutes per day, so the total minutes run by sixth graders is ( 20 times 3e = 60e ).- Seventh graders run an average of 18 minutes per day, so the total minutes run by seventh graders is ( 18 times 2e = 36e ).- Eighth graders run an average of 22 minutes per day, so the total minutes run by eighth graders is ( 22 times e = 22e ).Adding these together gives the total minutes run by all students:[60e + 36e + 22e = 118e]The total number of students is:[3e + 2e + e = 6e]To find the average number of minutes run per day by all students, we divide the total minutes by the total number of students:[frac{118e}{6e} = frac{118}{6} approx 19.67]Thus, the average number of minutes run per day by these students is approximately 19.67.