💡First, I need to determine the number of ways to travel from location A to location B. There are 3 train services available for this leg of the journey.Next, for each train service from A to B, there are 2 ferry services available to continue the journey from B to C.To find the total number of different travel options from A to C, I multiply the number of train services by the number of ferry services: 3 trains × 2 ferries = 6 travel options.Therefore, there are 6 different ways to travel from location A to location C.

💡Okay, so I have this problem about an ellipse. Let me try to figure it out step by step. First, part (1) asks for the standard equation of the ellipse E. The given equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b > 0). The sum of distances from any point on the ellipse to the two foci (F_1) and (F_2) is (2sqrt{3}), and the eccentricity is (frac{sqrt{3}}{3}). Hmm, I remember that for an ellipse, the sum of the distances from any point on the ellipse to the two foci is equal to (2a). So, if that sum is (2sqrt{3}), then (2a = 2sqrt{3}), which means (a = sqrt{3}). Got that.Next, the eccentricity (e) is given as (frac{sqrt{3}}{3}). Eccentricity is defined as (e = frac{c}{a}), where (c) is the distance from the center to each focus. So, plugging in the known values, we have:[e = frac{c}{a} implies frac{sqrt{3}}{3} = frac{c}{sqrt{3}} implies c = frac{sqrt{3} times sqrt{3}}{3} = frac{3}{3} = 1]So, (c = 1). Now, for an ellipse, we also know that (a^2 = b^2 + c^2). We have (a = sqrt{3}) and (c = 1), so let's plug those in:[a^2 = b^2 + c^2 implies (sqrt{3})^2 = b^2 + 1^2 implies 3 = b^2 + 1 implies b^2 = 3 - 1 = 2 implies b = sqrt{2}]So, (b = sqrt{2}). Therefore, the standard equation of the ellipse is:[frac{x^2}{(sqrt{3})^2} + frac{y^2}{(sqrt{2})^2} = 1 implies frac{x^2}{3} + frac{y^2}{2} = 1]Alright, that should be part (1). Let me just double-check my steps. Sum of distances is (2a), so (a = sqrt{3}). Eccentricity gives (c = 1). Then, using (a^2 = b^2 + c^2), we find (b^2 = 2). Yep, that seems right.Moving on to part (2). It says: Prove that the product of the slopes of line PQ and line OQ is a constant.Alright, so we have a moving point P on the line (x = 3). So, point P has coordinates ((3, y_0)), where (y_0) can be any real number. Then, a perpendicular line (l) to line (PF_2) through (F_2) intersects the ellipse at point Q. We need to find the product of the slopes of PQ and OQ and show it's a constant.First, let me visualize this. The ellipse is centered at the origin, right? Because the standard equation is (frac{x^2}{3} + frac{y^2}{2} = 1). So, the foci are located at ((pm c, 0)), which in this case is ((pm 1, 0)). So, (F_1 = (-1, 0)) and (F_2 = (1, 0)).Point P is on the line (x = 3), so it's somewhere along the vertical line three units to the right of the origin. Then, we draw a line from (F_2) (which is at (1, 0)) that is perpendicular to (PF_2). This line intersects the ellipse again at point Q. So, Q is another point on the ellipse, and we need to find the product of the slopes of PQ and OQ.Let me denote the coordinates:- (P = (3, y_0))- (F_2 = (1, 0))- (Q = (x_1, y_1))We need to find the slopes of PQ and OQ, then compute their product and show it's a constant.First, let's find the slope of (PF_2). The slope between (P(3, y_0)) and (F_2(1, 0)) is:[m_{PF_2} = frac{0 - y_0}{1 - 3} = frac{-y_0}{-2} = frac{y_0}{2}]Since line (l) is perpendicular to (PF_2), its slope will be the negative reciprocal of (m_{PF_2}). So, the slope of line (l) is:[m_l = -frac{2}{y_0}]But line (l) passes through (F_2(1, 0)), so its equation is:[y - 0 = m_l (x - 1) implies y = -frac{2}{y_0}(x - 1)]This line intersects the ellipse at point Q. So, to find Q, we can solve the system of equations:1. (frac{x^2}{3} + frac{y^2}{2} = 1)2. (y = -frac{2}{y_0}(x - 1))Let me substitute equation 2 into equation 1:[frac{x^2}{3} + frac{left(-frac{2}{y_0}(x - 1)right)^2}{2} = 1]Simplify this:First, square the y-term:[left(-frac{2}{y_0}(x - 1)right)^2 = frac{4}{y_0^2}(x - 1)^2]So, substituting back:[frac{x^2}{3} + frac{4}{2 y_0^2}(x - 1)^2 = 1 implies frac{x^2}{3} + frac{2}{y_0^2}(x - 1)^2 = 1]Multiply both sides by 3 (y_0^2) to eliminate denominators:[y_0^2 x^2 + 6(x - 1)^2 = 3 y_0^2]Expand ( (x - 1)^2 ):[(x - 1)^2 = x^2 - 2x + 1]So, substituting back:[y_0^2 x^2 + 6(x^2 - 2x + 1) = 3 y_0^2]Expand the terms:[y_0^2 x^2 + 6x^2 - 12x + 6 = 3 y_0^2]Bring all terms to one side:[y_0^2 x^2 + 6x^2 - 12x + 6 - 3 y_0^2 = 0]Factor terms:Group the (x^2) terms:[(y_0^2 + 6) x^2 - 12x + (6 - 3 y_0^2) = 0]This is a quadratic in x. Since we know that (F_2(1, 0)) is on both the ellipse and line (l), x = 1 is a root of this quadratic. So, we can factor (x - 1) out.Let me perform polynomial division or factorization. Let me denote the quadratic as:[A x^2 + B x + C = 0]Where:- (A = y_0^2 + 6)- (B = -12)- (C = 6 - 3 y_0^2)If x = 1 is a root, then plugging x = 1:[A(1)^2 + B(1) + C = A + B + C = (y_0^2 + 6) - 12 + (6 - 3 y_0^2) = y_0^2 + 6 - 12 + 6 - 3 y_0^2 = (-2 y_0^2) + 0 = -2 y_0^2]Wait, that's not zero unless (y_0 = 0), which isn't necessarily the case. Hmm, maybe I made a mistake.Wait, actually, (F_2(1, 0)) is on line (l), but is it on the ellipse? Let me check. Plug x = 1, y = 0 into the ellipse equation:[frac{1^2}{3} + frac{0^2}{2} = frac{1}{3} + 0 = frac{1}{3} neq 1]Oh! So, (F_2) is not on the ellipse. That's a mistake in my earlier reasoning. So, line (l) passes through (F_2) and intersects the ellipse at Q, but (F_2) is not on the ellipse. So, the quadratic equation we have will have two roots: one is the x-coordinate of Q, and the other is something else, but since (F_2) is not on the ellipse, we can't factor (x - 1) out.Hmm, so maybe I need another approach.Alternatively, since we know that (PF_2) is perpendicular to (F_2Q), the vectors (PF_2) and (F_2Q) are perpendicular. So, their dot product is zero.Let me write that down.Vector (PF_2) is from (P(3, y_0)) to (F_2(1, 0)):[overrightarrow{PF_2} = (1 - 3, 0 - y_0) = (-2, -y_0)]Vector (F_2Q) is from (F_2(1, 0)) to (Q(x_1, y_1)):[overrightarrow{F_2Q} = (x_1 - 1, y_1 - 0) = (x_1 - 1, y_1)]Since they are perpendicular:[overrightarrow{PF_2} cdot overrightarrow{F_2Q} = 0 implies (-2)(x_1 - 1) + (-y_0)(y_1) = 0]Simplify:[-2(x_1 - 1) - y_0 y_1 = 0 implies -2x_1 + 2 - y_0 y_1 = 0 implies -2x_1 - y_0 y_1 + 2 = 0]Let me write that as:[2x_1 + y_0 y_1 = 2]So, that's one equation involving (x_1) and (y_1).Also, since Q lies on the ellipse, it must satisfy the ellipse equation:[frac{x_1^2}{3} + frac{y_1^2}{2} = 1]So, we have two equations:1. (2x_1 + y_0 y_1 = 2)2. (frac{x_1^2}{3} + frac{y_1^2}{2} = 1)We need to find the slopes of PQ and OQ.First, let's find the slope of PQ. Points P(3, y_0) and Q(x_1, y_1):[m_{PQ} = frac{y_1 - y_0}{x_1 - 3}]Slope of OQ, where O is the origin (0, 0):[m_{OQ} = frac{y_1 - 0}{x_1 - 0} = frac{y_1}{x_1}]So, the product of the slopes is:[m_{PQ} times m_{OQ} = frac{y_1 - y_0}{x_1 - 3} times frac{y_1}{x_1} = frac{y_1(y_1 - y_0)}{x_1(x_1 - 3)}]We need to show that this product is a constant, independent of (y_0) or (x_1, y_1).Let me denote this product as K:[K = frac{y_1(y_1 - y_0)}{x_1(x_1 - 3)}]Our goal is to show that K is a constant.From equation 1: (2x_1 + y_0 y_1 = 2), we can solve for (y_0):[y_0 y_1 = 2 - 2x_1 implies y_0 = frac{2 - 2x_1}{y_1}]So, (y_0 = frac{2(1 - x_1)}{y_1})Let me substitute this into the expression for K:First, compute (y_1 - y_0):[y_1 - y_0 = y_1 - frac{2(1 - x_1)}{y_1} = frac{y_1^2 - 2(1 - x_1)}{y_1}]So, the numerator of K becomes:[y_1(y_1 - y_0) = y_1 times frac{y_1^2 - 2(1 - x_1)}{y_1} = y_1^2 - 2(1 - x_1)]Simplify:[y_1^2 - 2 + 2x_1]So, numerator is (y_1^2 + 2x_1 - 2)Denominator is (x_1(x_1 - 3))So, K becomes:[K = frac{y_1^2 + 2x_1 - 2}{x_1(x_1 - 3)}]Now, let's recall that Q lies on the ellipse, so:[frac{x_1^2}{3} + frac{y_1^2}{2} = 1 implies y_1^2 = 2left(1 - frac{x_1^2}{3}right) = 2 - frac{2x_1^2}{3}]So, substitute (y_1^2 = 2 - frac{2x_1^2}{3}) into the numerator:[y_1^2 + 2x_1 - 2 = left(2 - frac{2x_1^2}{3}right) + 2x_1 - 2 = -frac{2x_1^2}{3} + 2x_1]So, numerator simplifies to:[-frac{2x_1^2}{3} + 2x_1 = -frac{2x_1^2 - 6x_1}{3} = -frac{2x_1(x_1 - 3)}{3}]Therefore, K becomes:[K = frac{-frac{2x_1(x_1 - 3)}{3}}{x_1(x_1 - 3)} = frac{-2}{3}]So, K is equal to (-frac{2}{3}), which is a constant.Therefore, the product of the slopes of PQ and OQ is a constant, specifically (-frac{2}{3}).Let me recap to make sure I didn't make any mistakes:1. Found the standard equation of the ellipse using given sum of distances and eccentricity.2. For part (2), set up coordinates for points P, F2, and Q.3. Used the fact that vectors PF2 and F2Q are perpendicular to get an equation involving x1 and y1.4. Expressed y0 in terms of x1 and y1.5. Calculated the slopes of PQ and OQ, then their product.6. Substituted y1^2 from the ellipse equation into the product expression.7. Simplified and found that the product is a constant.Everything seems to check out. I think this is the correct approach.**Final Answer**(1) The standard equation of ellipse E is (boxed{dfrac{x^2}{3} + dfrac{y^2}{2} = 1}).(2) The product of the slopes of line PQ and line OQ is the constant (boxed{-dfrac{2}{3}}).

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. The problem says:"Given an arithmetic sequence {a_n} with a common difference of 2, if a₁, a₂, and a₄ form a geometric sequence, then a₁ equals (　). The options are A: 2, B: 1, C: -1, D: -2."Alright, so first, let me recall what an arithmetic sequence is. An arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference. In this case, the common difference is 2. So, that means:a₁ = a₁ (obviously)a₂ = a₁ + 2a₃ = a₁ + 4a₄ = a₁ + 6and so on.So, the nth term of an arithmetic sequence can be written as a_n = a₁ + (n - 1)d, where d is the common difference. Here, d = 2, so a_n = a₁ + 2(n - 1).Now, the problem says that a₁, a₂, and a₄ form a geometric sequence. Hmm, okay. Let me remember what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r.So, if a₁, a₂, a₄ are in a geometric sequence, then the ratio between a₂ and a₁ should be the same as the ratio between a₄ and a₂. In other words:a₂ / a₁ = a₄ / a₂This is the key property of a geometric sequence. So, I can set up this equation using the expressions for a₂ and a₄ from the arithmetic sequence.Let me write that out:a₂ / a₁ = a₄ / a₂Substituting the expressions from the arithmetic sequence:(a₁ + 2) / a₁ = (a₁ + 6) / (a₁ + 2)Okay, so now I have an equation in terms of a₁. Let me solve this equation to find the value of a₁.First, cross-multiplying to eliminate the fractions:(a₁ + 2)^2 = a₁(a₁ + 6)Let me expand both sides:Left side: (a₁ + 2)^2 = a₁^2 + 4a₁ + 4Right side: a₁(a₁ + 6) = a₁^2 + 6a₁So, putting it all together:a₁^2 + 4a₁ + 4 = a₁^2 + 6a₁Hmm, let me subtract a₁^2 from both sides to simplify:4a₁ + 4 = 6a₁Now, subtract 4a₁ from both sides:4 = 2a₁Divide both sides by 2:a₁ = 2Wait, so a₁ is 2? Let me check if this makes sense.If a₁ is 2, then the arithmetic sequence would be:a₁ = 2a₂ = 2 + 2 = 4a₃ = 4 + 2 = 6a₄ = 6 + 2 = 8So, the terms a₁, a₂, a₄ are 2, 4, 8.Is 2, 4, 8 a geometric sequence? Let's see.The ratio between a₂ and a₁ is 4 / 2 = 2.The ratio between a₄ and a₂ is 8 / 4 = 2.Yes, both ratios are equal to 2, so it is indeed a geometric sequence with a common ratio of 2.So, that seems to check out.But wait, let me make sure I didn't make any mistakes in my algebra. Let me go back through the steps.Starting with the equation:(a₁ + 2) / a₁ = (a₁ + 6) / (a₁ + 2)Cross-multiplying:(a₁ + 2)^2 = a₁(a₁ + 6)Expanding both sides:Left: a₁² + 4a₁ + 4Right: a₁² + 6a₁Subtracting a₁² from both sides:4a₁ + 4 = 6a₁Subtracting 4a₁:4 = 2a₁Divide by 2:a₁ = 2Yes, that seems correct.But wait, let me think again. The problem gives me options, and 2 is option A. But sometimes, when you square terms or cross-multiply, you might introduce extraneous solutions. Did I consider that?In this case, when I squared both sides, I didn't actually square both sides; I cross-multiplied, which is a valid step. So, I don't think there's an extraneous solution here. Let me plug a₁ = 2 back into the original equation to verify.Original equation:(a₁ + 2) / a₁ = (a₁ + 6) / (a₁ + 2)Plugging in a₁ = 2:(2 + 2) / 2 = (2 + 6) / (2 + 2)4 / 2 = 8 / 42 = 2Yes, that's correct. So, a₁ = 2 is a valid solution.But wait, the options include negative numbers as well. Could there be another solution?Let me think. When I solved the equation, I got a linear equation, which only has one solution. So, a₁ = 2 is the only solution. Therefore, the answer should be A: 2.But just to make sure, let me test another option, say, option D: -2.If a₁ = -2, then:a₁ = -2a₂ = -2 + 2 = 0a₄ = -2 + 6 = 4So, the terms would be -2, 0, 4.But in a geometric sequence, you can't have a zero term because the common ratio would be undefined (division by zero). So, that's not possible. So, a₁ can't be -2.Similarly, let me test option C: -1.a₁ = -1a₂ = -1 + 2 = 1a₄ = -1 + 6 = 5So, the terms would be -1, 1, 5.Is this a geometric sequence?Check the ratios:1 / (-1) = -15 / 1 = 5These ratios are not equal, so it's not a geometric sequence. So, a₁ can't be -1.How about option B: 1.a₁ = 1a₂ = 1 + 2 = 3a₄ = 1 + 6 = 7So, the terms would be 1, 3, 7.Check the ratios:3 / 1 = 37 / 3 ≈ 2.333...These are not equal, so it's not a geometric sequence. So, a₁ can't be 1.Therefore, the only valid solution is a₁ = 2, which is option A.Wait, but in the initial problem, the assistant's answer was also A: 2, but the user's initial problem had options A:2, B:1, C:-1, D:-2. So, the correct answer is A:2.But just to make sure, let me think if there's another way to approach this problem.Alternatively, since a₁, a₂, a₄ form a geometric sequence, we can denote them as:First term: a₁Second term: a₂ = a₁ + 2Third term: a₄ = a₁ + 6In a geometric sequence, the square of the second term is equal to the product of the first and third terms. So,(a₂)^2 = a₁ * a₄Which is the same as:(a₁ + 2)^2 = a₁(a₁ + 6)Which is exactly the equation I set up earlier. So, solving that gives a₁ = 2.Therefore, I'm confident that the correct answer is A:2.**Final Answer**boxed{A}

💡Okay, so I have this sequence {a_n} where a_1 is 1/2, and each subsequent term is defined by a_n = (2n - 3)/(2n) times a_{n-1}. I need to prove that the sum of the first n terms is less than 1 for any positive integer n. Hmm, let's see.First, maybe I should write out the first few terms to get a sense of what's happening. So, a_1 is 1/2. Then, a_2 would be (2*2 - 3)/(2*2) * a_1, which is (4 - 3)/4 * 1/2 = 1/4 * 1/2 = 1/8. Wait, no, that's not right. Let me recalculate that. (2*2 - 3) is 1, and 2*2 is 4, so 1/4 times a_1, which is 1/2, so 1/4 * 1/2 is 1/8. Okay, so a_2 is 1/8.Then, a_3 would be (2*3 - 3)/(2*3) * a_2. That's (6 - 3)/6 * 1/8 = 3/6 * 1/8 = 1/2 * 1/8 = 1/16. Hmm, interesting, so a_3 is 1/16.Wait, so a_1 is 1/2, a_2 is 1/8, a_3 is 1/16... Let me see if I can spot a pattern here. So, a_1 is 1/2, a_2 is 1/8, which is 1/(2^3), a_3 is 1/16, which is 1/(2^4). Hmm, maybe the pattern is that a_n is 1/(2^{n+1})? Let me check for a_4.a_4 would be (2*4 - 3)/(2*4) * a_3 = (8 - 3)/8 * 1/16 = 5/8 * 1/16 = 5/128. Wait, that's not 1/(2^5) which is 1/32. So, my initial pattern guess is wrong. Hmm.Maybe I need to find a general formula for a_n. Let's try to express a_n in terms of a_1. Since each term is a multiple of the previous term, it's a recursive sequence. So, a_n = (2n - 3)/(2n) * a_{n-1}.Let me write out the terms:a_1 = 1/2a_2 = (1/4) * a_1 = (1/4) * (1/2) = 1/8a_3 = (3/6) * a_2 = (1/2) * (1/8) = 1/16a_4 = (5/8) * a_3 = (5/8) * (1/16) = 5/128a_5 = (7/10) * a_4 = (7/10) * (5/128) = 35/1280 = 7/256Hmm, so a_1 = 1/2, a_2 = 1/8, a_3 = 1/16, a_4 = 5/128, a_5 = 7/256. It seems like the numerators are odd numbers: 1, 1, 1, 5, 7... Wait, not exactly. Let me see:a_1 numerator: 1a_2 numerator: 1a_3 numerator: 1a_4 numerator: 5a_5 numerator: 7Wait, maybe it's the product of odd numbers? Let's see:a_1 = 1/2a_2 = (1/4) * (1/2) = 1/(2*4) = 1/8a_3 = (3/6) * (1/8) = (3)/(6*8) = 3/48 = 1/16a_4 = (5/8) * (1/16) = 5/(8*16) = 5/128a_5 = (7/10) * (5/128) = 35/(10*128) = 35/1280 = 7/256So, the denominator seems to be 2^n * something. Wait, let me see:a_1: denominator 2a_2: denominator 8 = 2^3a_3: denominator 16 = 2^4a_4: denominator 128 = 2^7a_5: denominator 256 = 2^8Wait, that doesn't seem consistent. Maybe I'm approaching this the wrong way.Alternatively, perhaps I can express a_n as a product. Since each term is a multiple of the previous term, a_n = a_1 * product from k=2 to n of (2k - 3)/(2k).So, a_n = (1/2) * product from k=2 to n of (2k - 3)/(2k).Let me write that out:a_n = (1/2) * [(1/4) * (3/6) * (5/8) * ... * (2n - 3)/(2n)]Simplify each fraction:(1/4) = 1/4(3/6) = 1/2(5/8) = 5/8(7/10) = 7/10...(2n - 3)/(2n) = (2n - 3)/(2n)So, the product becomes:(1/4) * (1/2) * (5/8) * (7/10) * ... * (2n - 3)/(2n)Hmm, this seems complicated. Maybe I can write it in terms of factorials or double factorials.Wait, the numerator is 1 * 3 * 5 * ... * (2n - 3), which is the product of odd numbers up to (2n - 3). That's similar to a double factorial.Similarly, the denominator is 2 * 4 * 6 * ... * 2n, which is 2^n * n!.So, the product can be written as:[1 * 3 * 5 * ... * (2n - 3)] / [2^n * n!]And since 1 * 3 * 5 * ... * (2n - 3) is equal to (2n - 3)!!, where !! denotes the double factorial.So, a_n = (1/2) * [(2n - 3)!! / (2^n * n!)].Hmm, okay, so that's a general formula for a_n.Now, I need to find the sum from k=1 to n of a_k and show that it's less than 1.Let me denote S_n = sum_{k=1}^n a_k.So, S_n = a_1 + a_2 + ... + a_n.Given that a_1 = 1/2, and each subsequent a_k is positive and decreasing, since each term is multiplied by a factor less than 1.Wait, is each term decreasing? Let's check:a_1 = 1/2a_2 = 1/8a_3 = 1/16a_4 = 5/128 ≈ 0.0390625a_5 = 7/256 ≈ 0.02734375Yes, the terms are decreasing. So, the sequence is positive and decreasing.Therefore, the partial sums S_n are increasing, but we need to show they are bounded above by 1.So, maybe I can find an expression for S_n and then show that it's less than 1.Alternatively, perhaps I can find a telescoping series or relate it to a known series.Wait, let's think about the recurrence relation again.Given a_n = (2n - 3)/(2n) * a_{n-1}.Let me rearrange this:a_n / a_{n-1} = (2n - 3)/(2n)So, the ratio of consecutive terms is (2n - 3)/(2n).Hmm, perhaps I can write this as:a_n = a_{n-1} * (2n - 3)/(2n)Which can be rewritten as:a_n = a_{n-1} * (1 - 3/(2n))Hmm, not sure if that helps.Alternatively, maybe I can express a_n in terms of a_{n-1} and find a pattern.Wait, another approach: perhaps consider the sum S_n and try to find a recursive relation for S_n.Given that S_n = S_{n-1} + a_n.But since a_n = (2n - 3)/(2n) * a_{n-1}, maybe I can relate S_n to S_{n-1}.But I'm not sure if that's directly helpful.Alternatively, perhaps I can find a closed-form expression for S_n.Wait, considering that a_n is expressed in terms of double factorials, maybe I can relate the sum to some known series involving double factorials.Alternatively, perhaps I can use induction to prove that S_n < 1.Let me try that.Base case: n = 1.S_1 = a_1 = 1/2 < 1. True.Assume that for some k >= 1, S_k < 1.Then, S_{k+1} = S_k + a_{k+1}.Since a_{k+1} > 0, S_{k+1} > S_k.But we need to show that S_{k+1} < 1.But wait, if S_k < 1, and a_{k+1} is positive, then S_{k+1} = S_k + a_{k+1} < 1 + a_{k+1}.But we need to show that S_{k+1} < 1, which would require that a_{k+1} < 1 - S_k.But I don't know if that's necessarily true. Maybe induction isn't the best approach here.Alternatively, perhaps I can find an upper bound for each a_n and then sum those bounds.Given that a_n = (1/2) * [(2n - 3)!! / (2^n * n!)], maybe I can find an upper bound for a_n.Wait, let's see:(2n - 3)!! = 1 * 3 * 5 * ... * (2n - 3)And 2^n * n! = 2 * 4 * 6 * ... * 2nSo, the ratio is [1 * 3 * 5 * ... * (2n - 3)] / [2 * 4 * 6 * ... * 2n]Which can be written as product_{k=1}^{n-1} (2k - 1)/(2k + 2)Wait, maybe not. Alternatively, perhaps I can compare this to a known series.Wait, another idea: perhaps consider the sum S_n and relate it to an integral or use some inequality.Alternatively, perhaps I can find a telescoping product or sum.Wait, let's think about the ratio a_n / a_{n-1} = (2n - 3)/(2n).So, a_n = (2n - 3)/(2n) * a_{n-1}Let me write this as:a_n = a_{n-1} * (1 - 3/(2n))Hmm, so each term is a fraction of the previous term.Wait, maybe I can write a_n in terms of a_1:a_n = a_1 * product_{k=2}^n (2k - 3)/(2k)Which is:a_n = (1/2) * product_{k=2}^n (2k - 3)/(2k)Now, let's see if we can simplify this product.Let me write out the product:product_{k=2}^n (2k - 3)/(2k) = [1/4] * [3/6] * [5/8] * ... * [(2n - 3)/(2n)]Hmm, let's see if we can write this as:[1 * 3 * 5 * ... * (2n - 3)] / [4 * 6 * 8 * ... * 2n]Wait, the numerator is the product of odd numbers from 1 to (2n - 3), and the denominator is the product of even numbers from 4 to 2n.But 4 to 2n can be written as 2*2, 2*3, ..., 2*n.So, denominator = 2^{n-1} * (2 * 3 * ... * n) = 2^{n-1} * n! / 1!Wait, no, denominator is 4 * 6 * 8 * ... * 2n = 2*2 * 2*3 * 2*4 * ... * 2*n = 2^{n-1} * (2 * 3 * 4 * ... * n) = 2^{n-1} * (n! / 1!)Wait, actually, 4 * 6 * 8 * ... * 2n = 2^{n-1} * (2 * 3 * 4 * ... * n) = 2^{n-1} * (n! / 1!)Wait, but 2 * 3 * 4 * ... * n = n! / 1!Yes, so denominator = 2^{n-1} * (n! / 1!) = 2^{n-1} * n!Similarly, numerator is 1 * 3 * 5 * ... * (2n - 3). Let's see, how many terms are there?From k=2 to n, so n-1 terms.Each term is 2k - 3, where k goes from 2 to n.So, when k=2, term is 1; k=3, term is 3; k=4, term is 5; ...; k=n, term is 2n - 3.So, the numerator is the product of the first (n-1) odd numbers starting from 1.Which is equal to (2n - 3)!!, the double factorial.So, numerator = (2n - 3)!!Therefore, the product becomes:(2n - 3)!! / (2^{n-1} * n!)So, a_n = (1/2) * (2n - 3)!! / (2^{n-1} * n!) = (2n - 3)!! / (2^n * n!)Hmm, okay, so a_n = (2n - 3)!! / (2^n * n!)Now, I need to find S_n = sum_{k=1}^n a_k.So, S_n = sum_{k=1}^n (2k - 3)!! / (2^k * k!)Wait, but for k=1, (2*1 - 3)!! = (-1)!! which is undefined. Hmm, that's a problem.Wait, maybe my general formula is slightly off. Let me check for k=1.a_1 = 1/2.In the formula, (2*1 - 3)!! = (-1)!!, which is undefined. So, perhaps the formula is valid for k >= 2.So, maybe I should adjust the formula.Wait, let's see:For k=1, a_1 = 1/2.For k=2, a_2 = (1/4) * a_1 = 1/8.Using the formula: (2*2 - 3)!! / (2^2 * 2!) = (1)!! / (4 * 2) = 1 / 8. Correct.For k=3, a_3 = (3/6) * a_2 = 1/2 * 1/8 = 1/16.Formula: (2*3 - 3)!! / (2^3 * 3!) = (3)!! / (8 * 6) = 3 / 48 = 1/16. Correct.For k=4, a_4 = (5/8) * a_3 = 5/8 * 1/16 = 5/128.Formula: (2*4 - 3)!! / (2^4 * 4!) = (5)!! / (16 * 24) = (5*3*1) / 384 = 15 / 384 = 5 / 128. Correct.So, the formula works for k >= 2, but for k=1, it's undefined because (-1)!! is undefined.So, perhaps I can write a_n as:a_n = (2n - 3)!! / (2^n * n!) for n >= 2, and a_1 = 1/2.Okay, so now, S_n = a_1 + sum_{k=2}^n a_k = 1/2 + sum_{k=2}^n (2k - 3)!! / (2^k * k!)Hmm, not sure if that helps directly.Wait, maybe I can find a generating function for the sequence a_n.Alternatively, perhaps I can relate the sum S_n to an integral.Wait, another idea: perhaps consider the infinite series sum_{k=1}^infty a_k and show that it converges to a value less than 1.If I can show that the infinite series converges to something less than 1, then all partial sums S_n would be less than 1.So, let's consider the infinite series S = sum_{k=1}^infty a_k.Given that a_n = (2n - 3)!! / (2^n * n!) for n >= 2, and a_1 = 1/2.Wait, but for n=1, a_1 = 1/2.So, S = 1/2 + sum_{k=2}^infty (2k - 3)!! / (2^k * k!)Hmm, not sure if that's helpful.Wait, another approach: perhaps use induction with a better approach.Let me try induction again, but this time, let's assume that S_{n} < 1 - something, and then show that S_{n+1} < 1 - something else.Wait, let's see.Base case: n=1, S_1 = 1/2 < 1. True.Assume that S_k < 1 for some k >=1.Then, S_{k+1} = S_k + a_{k+1}.Since a_{k+1} > 0, S_{k+1} > S_k.But we need to show that S_{k+1} < 1.But from the assumption, S_k < 1, so S_{k+1} = S_k + a_{k+1} < 1 + a_{k+1}.But we need to show that S_{k+1} < 1, which would require that a_{k+1} < 1 - S_k.But I don't know if that's necessarily true. Maybe I need a different approach.Wait, perhaps I can find an upper bound for a_n.Given that a_n = (2n - 3)!! / (2^n * n!).I know that (2n - 3)!! = product_{k=1}^{n-1} (2k - 1).And n! = product_{k=1}^n k.So, a_n = [product_{k=1}^{n-1} (2k - 1)] / [2^n * product_{k=1}^n k]Hmm, maybe I can write this as:a_n = [product_{k=1}^{n-1} (2k - 1)] / [2^n * n!]Wait, perhaps I can relate this to the beta function or gamma function, but I'm not sure.Alternatively, perhaps I can use inequalities involving factorials.Wait, I recall that (2n)!! = 2^n * n!.Yes, because (2n)!! = 2 * 4 * 6 * ... * 2n = 2^n * (1 * 2 * 3 * ... * n) = 2^n * n!.Similarly, (2n - 1)!! = 1 * 3 * 5 * ... * (2n - 1).So, in our case, (2n - 3)!! = 1 * 3 * 5 * ... * (2n - 3).So, a_n = (2n - 3)!! / (2^n * n!) = [product_{k=1}^{n-1} (2k - 1)] / [2^n * n!]Hmm, maybe I can write this as:a_n = [ (2n - 2)! ) / (2^{n-1} * (n-1)! ) ] / (2^n * n! )Wait, no, that might not be correct.Wait, another idea: perhaps use the identity that (2n)!! * (2n - 1)!! = (2n)!.Yes, because (2n)! = 2n * (2n - 1) * 2(n - 1) * (2n - 3) * ... * 2 * 1 = (2n)!! * (2n - 1)!!.So, (2n)!! * (2n - 1)!! = (2n)!.Therefore, (2n - 1)!! = (2n)! / (2n)!! = (2n)! / (2^n * n!).So, (2n - 1)!! = (2n)! / (2^n * n!).Similarly, (2n - 3)!! = (2n - 2)! / (2^{n-1} * (n - 1)! )Wait, let me check:For (2n - 1)!!, we have (2n - 1)!! = (2n)! / (2^n * n!).Similarly, (2n - 3)!! = (2n - 2)! / (2^{n-1} * (n - 1)! )Yes, because (2n - 2)! = (2n - 2)!! * (2n - 3)!!.And (2n - 2)!! = 2^{n-1} * (n - 1)!.So, (2n - 3)!! = (2n - 2)! / (2^{n-1} * (n - 1)! )Therefore, a_n = (2n - 3)!! / (2^n * n!) = [ (2n - 2)! / (2^{n-1} * (n - 1)! ) ] / (2^n * n!) = (2n - 2)! / (2^{2n - 1} * (n - 1)! * n! )Hmm, that's a bit messy, but maybe I can simplify it further.Wait, let's see:(2n - 2)! / (n! * (n - 1)! ) = C(2n - 2, n - 1), which is the binomial coefficient.So, a_n = C(2n - 2, n - 1) / 2^{2n - 1}Hmm, interesting. So, a_n = C(2n - 2, n - 1) / 2^{2n - 1}Now, the binomial coefficient C(2n - 2, n - 1) is the number of ways to choose n - 1 items from 2n - 2, which is also equal to the central binomial coefficient divided by something.Wait, the central binomial coefficient is C(2n, n).But here, we have C(2n - 2, n - 1).I know that C(2n, n) ≈ 4^n / sqrt(π n) for large n, but I'm not sure if that helps here.Alternatively, perhaps I can use generating functions.Wait, the generating function for the central binomial coefficients is sum_{n=0}^infty C(2n, n) x^n = 1 / sqrt(1 - 4x).But in our case, we have C(2n - 2, n - 1) / 2^{2n - 1}.Hmm, maybe I can relate this to the generating function.Let me consider the generating function G(x) = sum_{n=1}^infty a_n x^n.Given that a_n = C(2n - 2, n - 1) / 2^{2n - 1}.So, G(x) = sum_{n=1}^infty [C(2n - 2, n - 1) / 2^{2n - 1}] x^n.Let me make a substitution: let m = n - 1. Then, when n=1, m=0, and G(x) becomes:sum_{m=0}^infty [C(2m, m) / 2^{2(m + 1) - 1}] x^{m + 1} = sum_{m=0}^infty [C(2m, m) / 2^{2m + 1}] x^{m + 1}.So, G(x) = (x / 2) * sum_{m=0}^infty [C(2m, m) / 4^m] x^m.But sum_{m=0}^infty C(2m, m) (x/4)^m = 1 / sqrt(1 - 4*(x/4)) = 1 / sqrt(1 - x).Therefore, G(x) = (x / 2) * 1 / sqrt(1 - x) = x / (2 sqrt(1 - x)).So, the generating function G(x) = x / (2 sqrt(1 - x)).Therefore, the sum S = sum_{n=1}^infty a_n = G(1).But wait, G(1) = 1 / (2 sqrt(1 - 1)) = 1 / 0, which is undefined. Hmm, that suggests that the series diverges, but that can't be right because the terms a_n are decreasing and positive, but perhaps the series converges.Wait, but actually, the generating function approach might not be directly applicable here because the radius of convergence is 1, and evaluating at x=1 might not be valid.Alternatively, perhaps I can use an integral representation.Wait, another idea: perhaps use the integral test for convergence.But since I'm dealing with a sum, maybe I can relate it to an integral.Alternatively, perhaps I can use the fact that a_n is related to the beta function.Wait, recalling that the beta function B(m, n) = integral_{0}^{1} t^{m-1} (1 - t)^{n-1} dt = (m-1)! (n-1)! / (m + n - 1)!.But I'm not sure if that helps here.Wait, another approach: perhaps consider the sum S_n and try to find a recursive relation.Given that a_n = (2n - 3)/(2n) * a_{n-1}, and S_n = S_{n-1} + a_n.So, S_n = S_{n-1} + (2n - 3)/(2n) * a_{n-1}.But a_{n-1} = (2n - 5)/(2n - 2) * a_{n-2}, and so on.Hmm, not sure if that helps.Wait, maybe I can write S_n in terms of S_{n-1} and a_{n-1}.But I'm stuck here.Wait, perhaps I can use the fact that a_n = (2n - 3)/(2n) * a_{n-1} and try to telescope the sum.Let me try to manipulate the recurrence relation.Given a_n = (2n - 3)/(2n) * a_{n-1}, let's rearrange:(2n) a_n = (2n - 3) a_{n-1}So, 2n a_n = (2n - 3) a_{n-1}Let me write this as:2n a_n - (2n - 3) a_{n-1} = 0Hmm, perhaps I can sum both sides from n=2 to n=k.Sum_{n=2}^k [2n a_n - (2n - 3) a_{n-1}] = 0So, Sum_{n=2}^k 2n a_n - Sum_{n=2}^k (2n - 3) a_{n-1} = 0Let me shift the index in the second sum:Let m = n - 1, so when n=2, m=1, and when n=k, m=k-1.So, Sum_{n=2}^k (2n - 3) a_{n-1} = Sum_{m=1}^{k-1} (2(m + 1) - 3) a_m = Sum_{m=1}^{k-1} (2m - 1) a_mTherefore, the equation becomes:Sum_{n=2}^k 2n a_n - Sum_{m=1}^{k-1} (2m - 1) a_m = 0Now, let's write both sums with the same index:Sum_{n=2}^k 2n a_n = Sum_{m=1}^{k-1} (2m - 1) a_mNow, let's express Sum_{n=2}^k 2n a_n as Sum_{n=1}^k 2n a_n - 2*1*a_1 = Sum_{n=1}^k 2n a_n - 2a_1Similarly, Sum_{m=1}^{k-1} (2m - 1) a_m = Sum_{m=1}^{k-1} 2m a_m - Sum_{m=1}^{k-1} a_mSo, putting it all together:Sum_{n=1}^k 2n a_n - 2a_1 = Sum_{m=1}^{k-1} 2m a_m - Sum_{m=1}^{k-1} a_mLet me denote Sum_{n=1}^k 2n a_n as 2 Sum_{n=1}^k n a_nSimilarly, Sum_{m=1}^{k-1} 2m a_m = 2 Sum_{m=1}^{k-1} m a_mAnd Sum_{m=1}^{k-1} a_m = S_{k-1}So, the equation becomes:2 Sum_{n=1}^k n a_n - 2a_1 = 2 Sum_{m=1}^{k-1} m a_m - S_{k-1}Let me rearrange terms:2 Sum_{n=1}^k n a_n - 2 Sum_{m=1}^{k-1} m a_m = 2a_1 - S_{k-1}The left side is 2 [Sum_{n=1}^k n a_n - Sum_{m=1}^{k-1} m a_m] = 2 [k a_k + Sum_{n=1}^{k-1} n a_n - Sum_{m=1}^{k-1} m a_m] = 2k a_kSo, 2k a_k = 2a_1 - S_{k-1}Therefore, S_{k-1} = 2a_1 - 2k a_kGiven that a_1 = 1/2, so 2a_1 = 1.Thus, S_{k-1} = 1 - 2k a_kTherefore, S_{k} = S_{k-1} + a_k = (1 - 2k a_k) + a_k = 1 - (2k - 1) a_kSo, S_k = 1 - (2k - 1) a_kTherefore, S_k = 1 - (2k - 1) a_kSince a_k > 0 for all k, then S_k < 1Because (2k - 1) a_k > 0, so 1 - (2k - 1) a_k < 1Therefore, S_k < 1Which is what we wanted to prove.So, the sum of the first n terms is less than 1 for any positive integer n.

💡Alright, so I have this geometry problem here, and I need to prove that AM equals AN in triangle ABC. Let me try to visualize the problem first. There's a triangle ABC, and AD is the angle bisector. They've constructed a circle using AD as its diameter, and this circle intersects sides AB and AC at points M and N, respectively, which are different from point A. I need to show that AM equals AN.Okay, let's start by drawing triangle ABC. Let me label the vertices A, B, and C. AD is the angle bisector, so it splits angle A into two equal parts. That means angle BAD is equal to angle CAD. Now, the circle is constructed with AD as its diameter. Hmm, circles with diameters have some special properties. I remember that any angle inscribed in a semicircle is a right angle. So, if I have a point on the circle, the angle subtended by the diameter at that point is 90 degrees.So, if the circle intersects AB at M and AC at N, then angles AMD and AND should be right angles. Let me write that down: angle AMD is 90 degrees, and angle AND is also 90 degrees. That makes sense because both M and N lie on the circle with diameter AD.Now, I have two right triangles: triangle AMD and triangle AND. Both share the side AD, which is the diameter of the circle. Since AD is common to both triangles, that might be useful. Also, both triangles have a right angle at M and N, respectively.I wonder if these two triangles are congruent. If they are, then their corresponding sides would be equal, which would mean AM equals AN. So, let's check the conditions for congruence. For two triangles to be congruent, we can use the Side-Angle-Side (SAS) criterion if two sides and the included angle are equal.In triangle AMD and triangle AND, we have:1. AD is common to both, so that's one side equal.2. Both have a right angle at M and N, so that's the included angle equal.3. But what about the other sides? We need to check if AM equals AN or if MD equals ND.Wait, I don't know if MD equals ND yet. Maybe I need another approach.Since AD is the angle bisector, angle BAD equals angle CAD. Also, both triangles AMD and AND have right angles at M and N. Maybe I can use some trigonometric relationships here.Let me consider triangle AMD first. Since angle AMD is 90 degrees, I can write that:sin(angle BAD) = opposite side / hypotenuse = AM / ADSimilarly, in triangle AND, angle AND is 90 degrees, so:sin(angle CAD) = opposite side / hypotenuse = AN / ADBut since angle BAD equals angle CAD (because AD is the angle bisector), their sines should be equal. Therefore, AM / AD equals AN / AD, which implies AM equals AN.Wait, that seems straightforward. So, using the sine of the equal angles and the common hypotenuse AD, we can conclude that AM equals AN.Alternatively, maybe I can use the properties of the circle and the angle bisector together. Since M and N lie on the circle with diameter AD, and AD is the angle bisector, there might be some symmetry involved.Let me think about the coordinates approach. Maybe assigning coordinates to the points could help. Let's place point A at the origin (0,0), point D somewhere along the x-axis, say at (d,0), and then points B and C somewhere in the plane. But this might get complicated, and I'm not sure if it's necessary.Going back to the right triangles idea, since both AMD and AND are right-angled and share the hypotenuse AD, and the angles at A are equal, then by the definition of sine, the opposite sides (AM and AN) must be equal.Yes, that makes sense. So, using the sine function in both triangles and the fact that the angles at A are equal, we can directly conclude that AM equals AN.I think that's a solid approach. It uses the properties of the circle, the angle bisector, and basic trigonometry to arrive at the conclusion. I don't see any flaws in this reasoning. Let me just recap to make sure I didn't miss anything.1. Circle with diameter AD implies angles AMD and AND are right angles.2. AD is the angle bisector, so angles BAD and CAD are equal.3. In right triangles AMD and AND, the sine of the equal angles at A gives AM / AD = AN / AD.4. Therefore, AM = AN.Yep, that seems to cover it. I think I've got it!

💡Okay, so I have this hyperbola problem here. The equation is given as ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ) with the condition that ( b > a > 0 ). I need to find the eccentricity of this hyperbola. The problem states that the symmetric point of the left focus about one of its asymptotes lies on the other asymptote. Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, I remember that for a hyperbola of the form ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ), the foci are located at ( (pm c, 0) ) where ( c = sqrt{a^2 + b^2} ). So, the left focus is at ( (-c, 0) ).Next, the asymptotes of this hyperbola are the lines that the hyperbola approaches but never touches. For this standard hyperbola, the asymptotes are ( y = pm frac{b}{a}x ). So, the two asymptotes are ( y = frac{b}{a}x ) and ( y = -frac{b}{a}x ).Now, the problem says that if we take the left focus, which is ( (-c, 0) ), and reflect it over one of the asymptotes, the resulting point lies on the other asymptote. Let me visualize this. If I reflect a point over a line, the reflected point should be such that the line is the perpendicular bisector of the segment joining the original point and its reflection.So, let's denote the left focus as ( F = (-c, 0) ). Let's pick one asymptote, say ( y = frac{b}{a}x ), and find the reflection of ( F ) over this line. Then, we need to check if this reflected point lies on the other asymptote ( y = -frac{b}{a}x ).To find the reflection of a point over a line, I can use the formula for reflection over a line. The general formula for reflecting a point ( (x, y) ) over the line ( ax + by + c = 0 ) is a bit involved, but since our asymptote is ( y = frac{b}{a}x ), which can be rewritten as ( frac{b}{a}x - y = 0 ), I can use this to find the reflection.Alternatively, I remember that reflecting a point over a line can be done using some vector projections. Let me recall the formula. If I have a point ( P ) and a line ( L ), the reflection ( P' ) of ( P ) over ( L ) can be found by:1. Finding the projection of ( P ) onto ( L ).2. Then, moving the same distance beyond the projection to get ( P' ).So, let's denote the asymptote as ( L: y = frac{b}{a}x ). Let me write this in standard form: ( frac{b}{a}x - y = 0 ).Given a point ( F = (-c, 0) ), I need to find its reflection over ( L ). Let me denote the reflection as ( F' ).First, I need to find the projection of ( F ) onto ( L ). The formula for the projection of a point ( (x_0, y_0) ) onto the line ( ax + by + c = 0 ) is:[left( frac{b(bx_0 - ay_0) - ac}{a^2 + b^2}, frac{a(-bx_0 + ay_0) - bc}{a^2 + b^2} right)]Wait, no, that doesn't seem right. Maybe I should use a different approach. Let me recall that the projection of a point ( P ) onto a line ( L ) can be found by solving the system where the line from ( P ) perpendicular to ( L ) intersects ( L ).So, the slope of ( L ) is ( frac{b}{a} ), so the slope of the perpendicular line is ( -frac{a}{b} ).Therefore, the equation of the line perpendicular to ( L ) passing through ( F = (-c, 0) ) is:[y - 0 = -frac{a}{b}(x + c)][y = -frac{a}{b}x - frac{a c}{b}]Now, the intersection point of this line with ( L ) is the projection of ( F ) onto ( L ). Let's find this intersection.We have two equations:1. ( y = frac{b}{a}x )2. ( y = -frac{a}{b}x - frac{a c}{b} )Set them equal:[frac{b}{a}x = -frac{a}{b}x - frac{a c}{b}]Multiply both sides by ( ab ) to eliminate denominators:[b^2 x = -a^2 x - a^2 c]Bring terms with ( x ) to one side:[b^2 x + a^2 x = -a^2 c][x(a^2 + b^2) = -a^2 c][x = -frac{a^2 c}{a^2 + b^2}]Now, substitute back into ( y = frac{b}{a}x ):[y = frac{b}{a} left( -frac{a^2 c}{a^2 + b^2} right) = -frac{a b c}{a^2 + b^2}]So, the projection point ( Q ) is ( left( -frac{a^2 c}{a^2 + b^2}, -frac{a b c}{a^2 + b^2} right) ).Now, to find the reflection ( F' ), we can use the fact that ( Q ) is the midpoint between ( F ) and ( F' ). So, if ( Q = left( frac{x_F + x_{F'}}{2}, frac{y_F + y_{F'}}{2} right) ), then:[-frac{a^2 c}{a^2 + b^2} = frac{-c + x_{F'}}{2}][-frac{a b c}{a^2 + b^2} = frac{0 + y_{F'}}{2}]Solving for ( x_{F'} ) and ( y_{F'} ):From the first equation:[- frac{a^2 c}{a^2 + b^2} = frac{-c + x_{F'}}{2}]Multiply both sides by 2:[- frac{2 a^2 c}{a^2 + b^2} = -c + x_{F'}]Bring ( -c ) to the left:[- frac{2 a^2 c}{a^2 + b^2} + c = x_{F'}]Factor out ( c ):[c left( 1 - frac{2 a^2}{a^2 + b^2} right) = x_{F'}]Simplify the expression inside the parentheses:[1 - frac{2 a^2}{a^2 + b^2} = frac{(a^2 + b^2) - 2 a^2}{a^2 + b^2} = frac{b^2 - a^2}{a^2 + b^2}]So,[x_{F'} = c cdot frac{b^2 - a^2}{a^2 + b^2} = frac{c (b^2 - a^2)}{a^2 + b^2}]From the second equation:[- frac{a b c}{a^2 + b^2} = frac{y_{F'}}{2}]Multiply both sides by 2:[y_{F'} = - frac{2 a b c}{a^2 + b^2}]So, the reflection point ( F' ) is ( left( frac{c (b^2 - a^2)}{a^2 + b^2}, - frac{2 a b c}{a^2 + b^2} right) ).Now, according to the problem, this reflected point ( F' ) lies on the other asymptote, which is ( y = -frac{b}{a}x ). So, let's plug the coordinates of ( F' ) into this equation and see what condition we get.So, substituting ( x = frac{c (b^2 - a^2)}{a^2 + b^2} ) and ( y = - frac{2 a b c}{a^2 + b^2} ) into ( y = -frac{b}{a}x ):Left side: ( y = - frac{2 a b c}{a^2 + b^2} )Right side: ( -frac{b}{a} cdot frac{c (b^2 - a^2)}{a^2 + b^2} = - frac{b c (b^2 - a^2)}{a(a^2 + b^2)} )Set them equal:[- frac{2 a b c}{a^2 + b^2} = - frac{b c (b^2 - a^2)}{a(a^2 + b^2)}]We can cancel out the negative signs and ( c ) from both sides (assuming ( c neq 0 ), which it isn't since ( b > a )):[frac{2 a b}{a^2 + b^2} = frac{b (b^2 - a^2)}{a(a^2 + b^2)}]Multiply both sides by ( a(a^2 + b^2) ) to eliminate denominators:[2 a b cdot a = b (b^2 - a^2)]Simplify left side:[2 a^2 b = b (b^2 - a^2)]Divide both sides by ( b ) (assuming ( b neq 0 )):[2 a^2 = b^2 - a^2]Bring ( a^2 ) to the left:[2 a^2 + a^2 = b^2][3 a^2 = b^2]So,[b = a sqrt{3}]Now, since ( c = sqrt{a^2 + b^2} ), substitute ( b^2 = 3 a^2 ):[c = sqrt{a^2 + 3 a^2} = sqrt{4 a^2} = 2 a]Therefore, the eccentricity ( e ) is given by ( e = frac{c}{a} = frac{2 a}{a} = 2 ).So, the eccentricity of the hyperbola is 2.Wait, let me just double-check my steps to make sure I didn't make any mistakes.1. Found the foci at ( (pm c, 0) ), correct.2. Asymptotes at ( y = pm frac{b}{a}x ), correct.3. Reflected the left focus over one asymptote and found the reflection point.4. Used the projection method to find the reflection, which seems correct.5. Set up the equation for the reflection point lying on the other asymptote, that seems right.6. Solved the equation and found ( b^2 = 3 a^2 ), which leads to ( c = 2 a ), so ( e = 2 ).Yes, everything seems consistent. I think the answer is correct.**Final Answer**The eccentricity of the hyperbola ( C ) is boxed{2}.