answer：First, I need to convert the waist size from inches to feet. Since there are 10 inches in a foot, I can divide the waist size in inches by 10 to get the measurement in feet.Next, I'll convert the waist size from feet to centimeters. Given that 1 foot is equal to 25 centimeters, I can multiply the waist size in feet by 25 to obtain the measurement in centimeters.Finally, I'll calculate the waist size in centimeters to determine the correct size the student should specify.

answer：First, I need to determine the marks obtained by each student based on the given percentages and relationships.Starting with D, who obtained 80% of the full marks of 500, I calculate D's marks as 0.80 multiplied by 500, which equals 400.Next, since C got 20% less than D, I find 20% of D's marks (which is 80) and subtract it from D's marks to get C's marks: 400 minus 80 equals 320.Then, B obtained 25% more than C. I calculate 25% of C's marks (which is 80) and add it to C's marks to find B's marks: 320 plus 80 equals 400.Finally, A got 10% less than B. I determine 10% of B's marks (which is 40) and subtract it from B's marks to find A's marks: 400 minus 40 equals 360.Therefore, A obtained 360 marks.

answer：Okay, so I need to figure out how many zeros are in the initial string of zeros after the decimal point in the decimal representation of ( frac{1}{20^{20}} ). The problem says that after these zeros, there's a 9 and then several more digits. The answer choices are from 23 to 27, so I need to be precise.First, let me understand what ( 20^{20} ) is. 20 is 2 times 10, so ( 20^{20} = (2 times 10)^{20} = 2^{20} times 10^{20} ). Therefore, ( frac{1}{20^{20}} = frac{1}{2^{20} times 10^{20}} ). I know that ( 2^{10} = 1024 ), so ( 2^{20} = (2^{10})^2 = 1024^2 ). Let me calculate that: 1024 times 1024. Hmm, 1000 times 1000 is 1,000,000, and then 24 times 1000 is 24,000, and 24 times 24 is 576. So adding those up: 1,000,000 + 24,000 + 24,000 + 576. Wait, that doesn't sound right. Maybe I should do it step by step.Actually, 1024 times 1024 is a known value. I think it's 1,048,576. Let me verify: 1024 times 1000 is 1,024,000, and 1024 times 24 is 24,576. Adding those together: 1,024,000 + 24,576 = 1,048,576. Yes, that's correct. So ( 2^{20} = 1,048,576 ).So, ( frac{1}{2^{20}} = frac{1}{1,048,576} ). I need to figure out what this is as a decimal. Since 1,048,576 is a little over 1,000,000, ( frac{1}{1,048,576} ) should be slightly less than ( frac{1}{1,000,000} = 0.000001 ). Let me see, ( frac{1}{1,048,576} ) is approximately 0.00000095367431640625. So, it's about 0.000000953674...So, in decimal form, it starts with six zeros after the decimal point, then a 9, and then more digits. So, ( frac{1}{2^{20}} ) is approximately 0.000000953674...Now, ( frac{1}{20^{20}} = frac{1}{2^{20} times 10^{20}} = frac{1}{2^{20}} times frac{1}{10^{20}} ). So, ( frac{1}{10^{20}} ) is 0.00000000000000000001, which is 1 followed by 20 zeros after the decimal point.Multiplying these two together: 0.000000953674... times 0.00000000000000000001. To multiply these, I can think of it as moving the decimal point. The first number has 6 zeros after the decimal before the 9, and the second number has 20 zeros. So, when I multiply them, the total number of zeros after the decimal before the first non-zero digit (which is 9) should be 6 + 20 = 26 zeros.Wait, but let me make sure. When multiplying two decimals, the number of decimal places is the sum of the decimal places of each number. The first number, ( frac{1}{2^{20}} ), is approximately 0.000000953674..., which has 6 zeros after the decimal before the 9. So, it's 7 decimal places in total (6 zeros + 1 digit). The second number, ( frac{1}{10^{20}} ), has 20 decimal places (all zeros except the last digit which is 1). So, multiplying these, the total number of decimal places is 7 + 20 = 27. But since the first number has 6 zeros before the 9, and the second number shifts it by 20 more places, the total number of zeros before the 9 would be 6 + 20 = 26.Let me think of it another way. If I have 0.000000953674... and I multiply it by 10^{-20}, which is the same as dividing by 10^{20}, it's like moving the decimal point 20 places to the left. So, starting from 0.000000953674..., moving the decimal 20 places to the left would add 20 more zeros after the decimal point before the 9. So, initially, there are 6 zeros, then after moving 20 places, it becomes 6 + 20 = 26 zeros before the 9.Therefore, the number of zeros in the initial string after the decimal point is 26.I think that's the answer. Let me just double-check. If I have ( frac{1}{20^{20}} ), which is ( frac{1}{(2 times 10)^{20}} = frac{1}{2^{20} times 10^{20}} ). We know ( 2^{20} ) is about a million, so ( frac{1}{2^{20}} ) is about a millionth, which is 0.000001, but slightly less, so 0.000000953674... Then, dividing that by ( 10^{20} ) moves the decimal 20 places to the right, which effectively adds 20 more zeros after the decimal before the 9. So, 6 zeros from the first part and 20 from the second part, totaling 26 zeros.Yeah, that makes sense. So, the answer should be 26 zeros.

answer：Okay, so I have this problem where I need to find the equation of a line that passes through the point (-1, 1) and intercepts a chord of length 4√3 on the circle given by the equation x² + y² - 2x - 4y - 11 = 0. Hmm, let me try to figure out how to approach this.First, I remember that the general equation of a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. So, I should probably rewrite the given circle equation into this standard form to find its center and radius.Starting with the given equation:x² + y² - 2x - 4y - 11 = 0.I can complete the squares for both x and y terms.For the x terms: x² - 2x. To complete the square, I take half of -2, which is -1, square it to get 1, and add and subtract it.For the y terms: y² - 4y. Similarly, half of -4 is -2, square it to get 4, and add and subtract it.So, rewriting the equation:(x² - 2x + 1 - 1) + (y² - 4y + 4 - 4) - 11 = 0.Simplify:(x - 1)² - 1 + (y - 2)² - 4 - 11 = 0.Combine constants:(x - 1)² + (y - 2)² - 1 - 4 - 11 = 0(x - 1)² + (y - 2)² - 16 = 0(x - 1)² + (y - 2)² = 16.So, the circle has center at (1, 2) and radius 4. Got that.Now, the problem is to find the equation of a line passing through (-1, 1) that intercepts a chord of length 4√3 on this circle.I remember that the length of a chord in a circle can be related to the distance from the center of the circle to the chord. The formula for the length of a chord is 2√(r² - d²), where r is the radius and d is the distance from the center to the chord.Given that the chord length is 4√3, so:4√3 = 2√(r² - d²)Divide both sides by 2:2√3 = √(r² - d²)Square both sides:(2√3)² = r² - d²4*3 = 16 - d²12 = 16 - d²d² = 16 - 12d² = 4d = 2.So, the distance from the center (1, 2) to the line must be 2 units.Now, the line passes through (-1, 1). Let me denote the equation of the line in point-slope form. Since it's a line, it can be written as y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point it passes through.So, substituting the point (-1, 1):y - 1 = m(x + 1)Which can be rewritten as:y = m(x + 1) + 1y = mx + m + 1Alternatively, in standard form: mx - y + (m + 1) = 0.Now, the distance from the center (1, 2) to this line should be 2. The formula for the distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / √(A² + B²).In our case, the line is mx - y + (m + 1) = 0, so A = m, B = -1, C = m + 1.So, the distance d is:|m*1 + (-1)*2 + (m + 1)| / √(m² + 1) = 2.Simplify the numerator:m - 2 + m + 1 = 2m - 1.So, |2m - 1| / √(m² + 1) = 2.So, we have |2m - 1| = 2√(m² + 1).Let me square both sides to eliminate the absolute value and the square root:(2m - 1)² = (2√(m² + 1))²4m² - 4m + 1 = 4(m² + 1)4m² - 4m + 1 = 4m² + 4Subtract 4m² from both sides:-4m + 1 = 4Subtract 1:-4m = 3Divide by -4:m = -3/4.So, the slope m is -3/4. Therefore, the equation of the line is:y - 1 = (-3/4)(x + 1)Let me write that in standard form:Multiply both sides by 4 to eliminate the fraction:4(y - 1) = -3(x + 1)4y - 4 = -3x - 3Bring all terms to one side:3x + 4y - 1 = 0.So, that's one possible line. But wait, when I squared both sides earlier, I might have introduced an extraneous solution. Let me check if this slope actually gives the correct distance.Compute |2m - 1| / √(m² + 1) with m = -3/4:|2*(-3/4) - 1| / √((9/16) + 1) = |-3/2 - 1| / √(25/16) = |-5/2| / (5/4) = (5/2) / (5/4) = (5/2)*(4/5) = 2. Okay, that checks out.But wait, earlier when I squared both sides, I might have missed the case where |2m - 1| could be negative. Let me think. The absolute value equation |2m - 1| = 2√(m² + 1) implies two cases:Case 1: 2m - 1 = 2√(m² + 1)Case 2: -(2m - 1) = 2√(m² + 1) => -2m + 1 = 2√(m² + 1)I only considered Case 1 earlier. Let me check Case 2.Case 2: -2m + 1 = 2√(m² + 1)Square both sides:(-2m + 1)² = 4(m² + 1)4m² - 4m + 1 = 4m² + 4Subtract 4m²:-4m + 1 = 4-4m = 3m = -3/4.Wait, that's the same solution as before. So, does that mean there's only one solution? Hmm, but sometimes when you square both sides, you might get the same solution for both cases, or sometimes no additional solutions.But let me think, could there be another line? Maybe a vertical line? Because sometimes lines can be vertical, which have undefined slopes, so they can't be expressed in slope-intercept form.Let me check if a vertical line passing through (-1, 1) could also satisfy the chord length condition.A vertical line has the equation x = -1. Let's see if this line intersects the circle and what chord length it would produce.Substitute x = -1 into the circle equation:(-1)² + y² - 2*(-1) - 4y - 11 = 01 + y² + 2 - 4y - 11 = 0y² - 4y - 8 = 0.Solve for y:y = [4 ± √(16 + 32)] / 2 = [4 ± √48] / 2 = [4 ± 4√3] / 2 = 2 ± 2√3.So, the two points of intersection are (-1, 2 + 2√3) and (-1, 2 - 2√3). The distance between these two points is:(2 + 2√3) - (2 - 2√3) = 4√3.So, the chord length is indeed 4√3. Therefore, the vertical line x = -1 is another solution.So, in total, there are two lines: one with slope -3/4, which is 3x + 4y - 1 = 0, and the vertical line x = -1.I think that's it. So, the equations of the lines are x = -1 and 3x + 4y - 1 = 0.

answer：Okay, so I have this problem about a circle and some projections and moving points. Let me try to break it down step by step. First, part (I). The circle is given by the equation (x^2 + y^2 = 4). So, it's a circle centered at the origin with radius 2. Point (P) is any point on this circle. The orthogonal projection of (P) onto the x-axis is point (Q). So, if (P) has coordinates ((x, y)), then (Q) must be ((x, 0)) because projecting onto the x-axis just drops the y-coordinate.Now, there's a moving point (M) such that the vector (overrightarrow{PQ}) is equal to twice the vector (overrightarrow{MQ}). Hmm, vectors can sometimes be tricky, but let's think about what this means. The vector (overrightarrow{PQ}) goes from (P) to (Q). Since (Q) is the projection of (P), this vector is just the vertical component from (P) to the x-axis. So, (overrightarrow{PQ}) would be ((0, -y)). Similarly, (overrightarrow{MQ}) goes from (M) to (Q). If (M) has coordinates ((a, b)), then (overrightarrow{MQ}) is ((x - a, 0 - b)).According to the problem, (overrightarrow{PQ} = 2overrightarrow{MQ}). So, substituting the vectors we have:[(0, -y) = 2(x - a, -b)]Breaking this into components:1. For the x-component: (0 = 2(x - a))2. For the y-component: (-y = 2(-b))From the x-component equation: (0 = 2(x - a)) implies (x - a = 0), so (a = x).From the y-component equation: (-y = -2b) implies (y = 2b), so (b = frac{y}{2}).Therefore, the coordinates of (M) are ((x, frac{y}{2})). But (P) is on the circle (x^2 + y^2 = 4). So, substituting (x) and (y) in terms of (M)'s coordinates, since (x = a) and (y = 2b), we have:[a^2 + (2b)^2 = 4][a^2 + 4b^2 = 4]So, replacing (a) and (b) with (x) and (y) (since (M) is ((x, y))), the equation becomes:[x^2 + 4y^2 = 4]Dividing both sides by 4 to simplify:[frac{x^2}{4} + y^2 = 1]So, that's the equation of curve (C). It looks like an ellipse with semi-major axis 2 along the x-axis and semi-minor axis 1 along the y-axis. That makes sense because the projection and scaling would transform the circle into an ellipse.Alright, that was part (I). Now, moving on to part (II). Point (A) is given as ((2, 0)), which is on curve (C). A line (l) passes through the point ((1, 0)) and intersects curve (C) at points (B) and (D). We need to find the slopes (k_1) and (k_2) of lines (AB) and (AD), respectively, and show that the product (k_1k_2) is a constant.Hmm, okay. Let's think about how to approach this. Since line (l) passes through ((1, 0)), we can write its equation in some form. Maybe parametric or slope-intercept. Let's consider both possibilities.First, let's denote the equation of line (l). Since it passes through ((1, 0)), if we let the slope be (m), the equation would be (y = m(x - 1)). Alternatively, if the line is vertical, the equation would be (x = 1). But let's check if (x = 1) intersects curve (C).Substituting (x = 1) into (C)'s equation:[frac{1^2}{4} + y^2 = 1 implies frac{1}{4} + y^2 = 1 implies y^2 = frac{3}{4} implies y = pm frac{sqrt{3}}{2}]So, points (B) and (D) would be ((1, frac{sqrt{3}}{2})) and ((1, -frac{sqrt{3}}{2})). Then, lines (AB) and (AD) would have slopes:For (AB): from (A(2, 0)) to (B(1, frac{sqrt{3}}{2})):[k_1 = frac{frac{sqrt{3}}{2} - 0}{1 - 2} = frac{frac{sqrt{3}}{2}}{-1} = -frac{sqrt{3}}{2}]For (AD): from (A(2, 0)) to (D(1, -frac{sqrt{3}}{2})):[k_2 = frac{-frac{sqrt{3}}{2} - 0}{1 - 2} = frac{-frac{sqrt{3}}{2}}{-1} = frac{sqrt{3}}{2}]Multiplying (k_1) and (k_2):[k_1k_2 = left(-frac{sqrt{3}}{2}right)left(frac{sqrt{3}}{2}right) = -frac{3}{4}]So, in this case, the product is (-frac{3}{4}). Interesting. Now, let's see if this holds for a general line passing through ((1, 0)).Let's consider a general line with slope (m), so its equation is (y = m(x - 1)). We need to find the points (B) and (D) where this line intersects curve (C). Substituting (y = m(x - 1)) into (C)'s equation:[frac{x^2}{4} + [m(x - 1)]^2 = 1][frac{x^2}{4} + m^2(x^2 - 2x + 1) = 1]Multiply through by 4 to eliminate the denominator:[x^2 + 4m^2(x^2 - 2x + 1) = 4][x^2 + 4m^2x^2 - 8m^2x + 4m^2 = 4][(1 + 4m^2)x^2 - 8m^2x + (4m^2 - 4) = 0]This is a quadratic in (x). Let's denote it as:[Ax^2 + Bx + C = 0]where:- (A = 1 + 4m^2)- (B = -8m^2)- (C = 4m^2 - 4)Let the roots be (x_1) and (x_2), corresponding to points (B) and (D). From Vieta's formulas, we know:[x_1 + x_2 = frac{-B}{A} = frac{8m^2}{1 + 4m^2}][x_1x_2 = frac{C}{A} = frac{4m^2 - 4}{1 + 4m^2}]Now, the corresponding (y)-coordinates for (B) and (D) are (y_1 = m(x_1 - 1)) and (y_2 = m(x_2 - 1)).We need to find the slopes (k_1) and (k_2). Slope (k_1) is the slope of line (AB), which connects (A(2, 0)) and (B(x_1, y_1)):[k_1 = frac{y_1 - 0}{x_1 - 2} = frac{m(x_1 - 1)}{x_1 - 2}]Similarly, slope (k_2) is the slope of line (AD), connecting (A(2, 0)) and (D(x_2, y_2)):[k_2 = frac{y_2 - 0}{x_2 - 2} = frac{m(x_2 - 1)}{x_2 - 2}]So, the product (k_1k_2) is:[k_1k_2 = left(frac{m(x_1 - 1)}{x_1 - 2}right)left(frac{m(x_2 - 1)}{x_2 - 2}right) = m^2 cdot frac{(x_1 - 1)(x_2 - 1)}{(x_1 - 2)(x_2 - 2)}]Let's compute the numerator and denominator separately.First, numerator: ((x_1 - 1)(x_2 - 1))Expanding this:[x_1x_2 - x_1 - x_2 + 1]From Vieta's formulas, we have (x_1 + x_2 = frac{8m^2}{1 + 4m^2}) and (x_1x_2 = frac{4m^2 - 4}{1 + 4m^2}). Plugging these in:[frac{4m^2 - 4}{1 + 4m^2} - frac{8m^2}{1 + 4m^2} + 1][= frac{4m^2 - 4 - 8m^2 + (1 + 4m^2)}{1 + 4m^2}]Wait, let me compute step by step:First term: (x_1x_2 = frac{4m^2 - 4}{1 + 4m^2})Second term: (-x_1 - x_2 = -frac{8m^2}{1 + 4m^2})Third term: (+1)So, altogether:[frac{4m^2 - 4}{1 + 4m^2} - frac{8m^2}{1 + 4m^2} + 1]Combine the fractions:[frac{4m^2 - 4 - 8m^2}{1 + 4m^2} + 1 = frac{-4m^2 - 4}{1 + 4m^2} + 1][= frac{-4(m^2 + 1)}{1 + 4m^2} + 1]Convert 1 to have the same denominator:[= frac{-4(m^2 + 1)}{1 + 4m^2} + frac{1 + 4m^2}{1 + 4m^2}][= frac{-4m^2 - 4 + 1 + 4m^2}{1 + 4m^2}]Simplify numerator:[(-4m^2 + 4m^2) + (-4 + 1) = 0 - 3 = -3]So, numerator is (-3).Now, the denominator: ((x_1 - 2)(x_2 - 2))Expanding this:[x_1x_2 - 2x_1 - 2x_2 + 4]Again, using Vieta's formulas:[x_1x_2 = frac{4m^2 - 4}{1 + 4m^2}][-2x_1 - 2x_2 = -2(x_1 + x_2) = -2 cdot frac{8m^2}{1 + 4m^2} = frac{-16m^2}{1 + 4m^2}][+4]So, altogether:[frac{4m^2 - 4}{1 + 4m^2} + frac{-16m^2}{1 + 4m^2} + 4]Combine the fractions:[frac{4m^2 - 4 - 16m^2}{1 + 4m^2} + 4 = frac{-12m^2 - 4}{1 + 4m^2} + 4]Convert 4 to have the same denominator:[= frac{-12m^2 - 4}{1 + 4m^2} + frac{4(1 + 4m^2)}{1 + 4m^2}][= frac{-12m^2 - 4 + 4 + 16m^2}{1 + 4m^2}]Simplify numerator:[(-12m^2 + 16m^2) + (-4 + 4) = 4m^2 + 0 = 4m^2]So, denominator is (frac{4m^2}{1 + 4m^2}).Putting it all together, the product (k_1k_2) is:[k_1k_2 = m^2 cdot frac{-3}{frac{4m^2}{1 + 4m^2}} = m^2 cdot left( frac{-3(1 + 4m^2)}{4m^2} right ) = frac{-3(1 + 4m^2)}{4}]Wait, hold on. Let me check that step again.Wait, no. The numerator was (-3) and the denominator was (frac{4m^2}{1 + 4m^2}). So, the fraction is:[frac{-3}{frac{4m^2}{1 + 4m^2}} = -3 cdot frac{1 + 4m^2}{4m^2} = frac{-3(1 + 4m^2)}{4m^2}]Therefore, (k_1k_2 = m^2 cdot frac{-3(1 + 4m^2)}{4m^2}). The (m^2) cancels out:[k_1k_2 = frac{-3(1 + 4m^2)}{4}]Wait, but that still has (m) in it. Hmm, that can't be right because we were supposed to get a constant. Did I make a mistake somewhere?Let me go back. Maybe I messed up the calculation when expanding the numerator or denominator.Wait, let's recast the problem. Maybe instead of parameterizing by slope (m), I can parameterize the line differently. Let me try using a parameter (t) such that the line is expressed as (x = ty + 1). Sometimes, using a different parameterization can simplify things.So, let me set the line (l) as (x = ty + 1), where (t) is a parameter. Then, substitute this into the equation of curve (C):[frac{(ty + 1)^2}{4} + y^2 = 1]Expanding:[frac{t^2y^2 + 2ty + 1}{4} + y^2 = 1]Multiply through by 4:[t^2y^2 + 2ty + 1 + 4y^2 = 4]Combine like terms:[(t^2 + 4)y^2 + 2ty + (1 - 4) = 0][(t^2 + 4)y^2 + 2ty - 3 = 0]This is a quadratic in (y). Let the roots be (y_1) and (y_2). Then, from Vieta's formulas:[y_1 + y_2 = frac{-2t}{t^2 + 4}][y_1y_2 = frac{-3}{t^2 + 4}]Now, the points (B) and (D) have coordinates ((tx_1 + 1, y_1)) and ((tx_2 + 1, y_2)). Wait, no. Wait, the line is (x = ty + 1), so for each (y), (x = ty + 1). So, if (y_1) and (y_2) are the y-coordinates, then the x-coordinates are (x_1 = ty_1 + 1) and (x_2 = ty_2 + 1).So, points (B) and (D) are ((ty_1 + 1, y_1)) and ((ty_2 + 1, y_2)).Now, let's compute the slopes (k_1) and (k_2).Slope (k_1) is the slope of line (AB), connecting (A(2, 0)) and (B(ty_1 + 1, y_1)):[k_1 = frac{y_1 - 0}{(ty_1 + 1) - 2} = frac{y_1}{ty_1 - 1}]Similarly, slope (k_2) is the slope of line (AD), connecting (A(2, 0)) and (D(ty_2 + 1, y_2)):[k_2 = frac{y_2 - 0}{(ty_2 + 1) - 2} = frac{y_2}{ty_2 - 1}]So, the product (k_1k_2) is:[k_1k_2 = left(frac{y_1}{ty_1 - 1}right)left(frac{y_2}{ty_2 - 1}right) = frac{y_1y_2}{(ty_1 - 1)(ty_2 - 1)}]Let's compute the numerator and denominator.Numerator: (y_1y_2 = frac{-3}{t^2 + 4}) from Vieta.Denominator: ((ty_1 - 1)(ty_2 - 1))Expanding this:[t^2y_1y_2 - t(y_1 + y_2) + 1]Substituting from Vieta's formulas:[t^2 cdot left(frac{-3}{t^2 + 4}right) - t cdot left(frac{-2t}{t^2 + 4}right) + 1]Simplify each term:First term: (frac{-3t^2}{t^2 + 4})Second term: (frac{2t^2}{t^2 + 4})Third term: (1)Combine them:[frac{-3t^2 + 2t^2}{t^2 + 4} + 1 = frac{-t^2}{t^2 + 4} + 1]Convert 1 to have the same denominator:[= frac{-t^2}{t^2 + 4} + frac{t^2 + 4}{t^2 + 4} = frac{-t^2 + t^2 + 4}{t^2 + 4} = frac{4}{t^2 + 4}]So, denominator is (frac{4}{t^2 + 4}).Therefore, the product (k_1k_2) is:[k_1k_2 = frac{frac{-3}{t^2 + 4}}{frac{4}{t^2 + 4}} = frac{-3}{4}]Ah, there we go! So, regardless of the parameter (t), the product (k_1k_2) is (-frac{3}{4}), which is a constant. That confirms the result we saw earlier when the line was vertical.So, whether the line is vertical or has some slope (m), the product of the slopes (k_1) and (k_2) is always (-frac{3}{4}). Therefore, it's proven that (k_1k_2) is a constant.I think that's a solid approach. By parameterizing the line differently, I was able to avoid the confusion I had earlier and clearly see that the product is indeed a constant. It's interesting how changing the parameterization can sometimes simplify the problem.Another thing I noticed is that curve (C) is an ellipse, and point (A) is one of its vertices. The line passing through ((1, 0)) seems to have a special property with respect to point (A), resulting in the constant product of slopes. It might be related to some geometric property of ellipses, perhaps related to their foci or something else, but I'm not sure. Regardless, the algebra checks out, so the conclusion is valid.In summary, for part (I), the curve (C) is an ellipse given by (frac{x^2}{4} + y^2 = 1). For part (II), regardless of the line passing through ((1, 0)), the product of the slopes (k_1) and (k_2) is always (-frac{3}{4}), which is a constant.**Final Answer**Ⅰ. The equation of curve (C) is boxed{dfrac{x^2}{4} + y^2 = 1}.Ⅱ. The constant value of (k_1k_2) is boxed{-dfrac{3}{4}}.

answer：Okay, so I need to solve the equation ( x + sqrt{2x - 3} = 5 ). Hmm, let me think about how to approach this. I remember that when dealing with equations that have square roots, a common strategy is to isolate the square root term and then square both sides to eliminate it. That should help me get rid of the square root and turn the equation into something more manageable, like a quadratic equation.Alright, let's start by isolating the square root. I'll subtract ( x ) from both sides of the equation:[ sqrt{2x - 3} = 5 - x ]Okay, now the square root is by itself on the left side. The next step is to square both sides to get rid of the square root. So, squaring the left side gives me ( 2x - 3 ), and squaring the right side gives me ( (5 - x)^2 ). Let me write that out:[ (2x - 3) = (5 - x)^2 ]Now, I need to expand the right side. I remember that ( (a - b)^2 = a^2 - 2ab + b^2 ), so applying that here:[ (5 - x)^2 = 25 - 10x + x^2 ]So now, the equation becomes:[ 2x - 3 = 25 - 10x + x^2 ]Hmm, this looks like a quadratic equation. To solve it, I should bring all terms to one side so that the equation equals zero. Let me subtract ( 2x - 3 ) from both sides:[ 0 = 25 - 10x + x^2 - 2x + 3 ]Simplifying the right side by combining like terms:First, combine the constant terms: ( 25 + 3 = 28 )Next, combine the ( x ) terms: ( -10x - 2x = -12x )So, the equation becomes:[ 0 = x^2 - 12x + 28 ]Alternatively, I can write it as:[ x^2 - 12x + 28 = 0 ]Now, I need to solve this quadratic equation. I can try factoring it. Let me look for two numbers that multiply to 28 and add up to -12. Hmm, factors of 28 are 1 & 28, 2 & 14, 4 & 7. Let's see:-4 and -7 multiply to 28 and add up to -11. That's not quite -12.Wait, maybe I made a mistake. Let me check again. 28 is positive, and the middle term is -12x, so both numbers should be negative because their product is positive and their sum is negative.So, -4 and -7 multiply to 28 and add up to -11. Hmm, not -12. Maybe it's not factorable easily. Let me try another approach.Alternatively, I can use the quadratic formula. The quadratic formula is ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -12 ), and ( c = 28 ).Plugging in the values:[ x = frac{-(-12) pm sqrt{(-12)^2 - 4 cdot 1 cdot 28}}{2 cdot 1} ][ x = frac{12 pm sqrt{144 - 112}}{2} ][ x = frac{12 pm sqrt{32}}{2} ][ x = frac{12 pm 4sqrt{2}}{2} ][ x = 6 pm 2sqrt{2} ]Wait, so the solutions are ( x = 6 + 2sqrt{2} ) and ( x = 6 - 2sqrt{2} ). Let me approximate these to see what they are approximately equal to.( sqrt{2} ) is approximately 1.414, so:( 2sqrt{2} approx 2.828 )So,( x = 6 + 2.828 approx 8.828 )and( x = 6 - 2.828 approx 3.172 )Hmm, so I have two potential solutions: approximately 8.828 and 3.172. But wait, I need to check these solutions in the original equation because sometimes squaring both sides can introduce extraneous solutions.Let me check ( x = 8.828 ) first.Original equation:[ x + sqrt{2x - 3} = 5 ]Plugging in ( x = 8.828 ):Left side: ( 8.828 + sqrt{2 cdot 8.828 - 3} )Calculate inside the square root:( 2 cdot 8.828 = 17.656 )( 17.656 - 3 = 14.656 )So, square root of 14.656 is approximately 3.828.So, left side: ( 8.828 + 3.828 approx 12.656 ), which is not equal to 5. So, this solution is extraneous.Now, let's check ( x = 3.172 ).Left side: ( 3.172 + sqrt{2 cdot 3.172 - 3} )Calculate inside the square root:( 2 cdot 3.172 = 6.344 )( 6.344 - 3 = 3.344 )Square root of 3.344 is approximately 1.828.So, left side: ( 3.172 + 1.828 = 5 ), which matches the right side.So, ( x = 3.172 ) is a valid solution.Wait, but earlier I thought I had two solutions, but one was extraneous. So, does that mean there's only one real solution?But hold on, let me think again. When I squared both sides, I might have introduced an extraneous solution, but in this case, only one of the two solutions worked. So, does that mean there's only one real root?But wait, looking back at the quadratic equation, I had two solutions, but only one worked when plugged back into the original equation. So, the equation has one real root.But let me double-check my calculations because sometimes I might make a mistake.Wait, when I squared both sides, I got ( x^2 - 12x + 28 = 0 ), which I solved using the quadratic formula and got ( x = 6 pm 2sqrt{2} ). But when I checked, only ( x = 6 - 2sqrt{2} ) worked, which is approximately 3.172, and ( x = 6 + 2sqrt{2} ) did not work.So, that suggests that the equation has only one real root.But wait, looking back at the original equation, ( x + sqrt{2x - 3} = 5 ), the domain of the square root function ( sqrt{2x - 3} ) requires that ( 2x - 3 geq 0 ), so ( x geq 1.5 ). So, both solutions ( x = 6 + 2sqrt{2} approx 8.828 ) and ( x = 6 - 2sqrt{2} approx 3.172 ) are greater than 1.5, so they are within the domain.But when I plugged in ( x = 8.828 ), it didn't satisfy the original equation, which is strange because both solutions are within the domain. Maybe I made a mistake in my calculations when checking.Let me recalculate ( x = 8.828 ):Left side: ( 8.828 + sqrt{2 cdot 8.828 - 3} )Calculate inside the square root:( 2 cdot 8.828 = 17.656 )( 17.656 - 3 = 14.656 )Square root of 14.656 is approximately 3.828.So, left side: ( 8.828 + 3.828 = 12.656 ), which is not equal to 5. So, it's definitely not a solution.Wait, but why did squaring both sides give me an extraneous solution? Because squaring can sometimes introduce solutions that don't satisfy the original equation, especially when dealing with square roots.So, in this case, even though both solutions are within the domain, only one satisfies the original equation. Therefore, the equation has only one real root.But wait, the options given are:(A) 2 real roots(B) 1 real and 1 imaginary root(C) 2 imaginary roots(D) no roots(E) 1 real rootSo, based on my calculations, the equation has one real root, which is ( x = 6 - 2sqrt{2} ). Therefore, the correct answer should be (E) 1 real root.But wait, let me think again. When I solved the quadratic equation, I got two real roots, but one was extraneous. So, does that mean the original equation has only one real root? Yes, because the other solution doesn't satisfy the original equation.Therefore, the answer is (E) 1 real root.