answer：Okay, so I have this inequality to solve: (frac{x - 3}{x^2 - 4x + 13} ge 0). Hmm, let me think about how to approach this. I remember that to solve rational inequalities, I need to find where the expression is positive or zero. That means I have to look at both the numerator and the denominator.First, let me look at the numerator: (x - 3). This is a linear expression, and it equals zero when (x = 3). So, that's a critical point. Now, the denominator is a quadratic: (x^2 - 4x + 13). I need to check if this quadratic ever equals zero because that would make the expression undefined, and those points are important for determining intervals.To find the roots of the quadratic, I can use the quadratic formula: (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}). Here, (a = 1), (b = -4), and (c = 13). Plugging those in: (x = frac{4 pm sqrt{(-4)^2 - 4*1*13}}{2*1}). Calculating the discriminant: (16 - 52 = -36). Oh, the discriminant is negative, which means there are no real roots. So, the quadratic doesn't cross the x-axis and is always positive or always negative.Since the coefficient of (x^2) is positive (1), the parabola opens upwards, meaning the quadratic is always positive. So, the denominator is always positive for all real numbers (x). That simplifies things because the sign of the entire expression depends solely on the numerator.The numerator (x - 3) is zero at (x = 3), positive when (x > 3), and negative when (x < 3). Since the denominator is always positive, the inequality (frac{x - 3}{x^2 - 4x + 13} ge 0) will hold true when the numerator is non-negative. That is, when (x - 3 ge 0), which simplifies to (x ge 3).So, the solution is all real numbers greater than or equal to 3. In interval notation, that's ([3, infty)).Wait, let me double-check. The denominator is always positive, so the sign of the expression is determined by the numerator. When (x = 3), the expression is zero, which satisfies the inequality. For (x > 3), the numerator is positive, so the expression is positive, which also satisfies the inequality. For (x < 3), the numerator is negative, so the expression is negative, which doesn't satisfy the inequality. Therefore, the solution is indeed (x ge 3).I think that's it. No mistakes noticed this time.

answer：First, I need to simplify the complex fraction ( z = frac{1 + 5i}{5 - i} ).To eliminate the imaginary unit in the denominator, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is ( 5 + i ).This gives me:[z = frac{(1 + 5i)(5 + i)}{(5 - i)(5 + i)}]Next, I'll expand both the numerator and the denominator.For the numerator:[(1 + 5i)(5 + i) = 1 cdot 5 + 1 cdot i + 5i cdot 5 + 5i cdot i = 5 + i + 25i + 5i^2]Since ( i^2 = -1 ), this simplifies to:[5 + 26i - 5 = 26i]For the denominator:[(5 - i)(5 + i) = 5 cdot 5 + 5 cdot i - i cdot 5 - i cdot i = 25 + 5i - 5i - i^2]Again, since ( i^2 = -1 ), this simplifies to:[25 + 0i + 1 = 26]Now, the fraction becomes:[z = frac{26i}{26} = i]Therefore, the correct answer is ( i ), which corresponds to option C.

answer：Okay, so I have this problem here about functions that satisfy the equation f(x + 6) + f(x - 6) = f(x) for all real x. I need to find the least common positive period p for all such functions. The options given are 12, 18, 24, 36, and 72. Hmm, interesting.First, I remember that functional equations often involve some kind of periodicity or recurrence relations. Since the equation relates f(x + 6) and f(x - 6) to f(x), it suggests that the function has some periodic behavior with a period related to 6. But I need to find the least common period, so maybe it's a multiple of 6?Let me try to manipulate the given equation to see if I can find a recurrence relation or something that can help me figure out the period. The equation is:f(x + 6) + f(x - 6) = f(x)Maybe I can substitute x with x + 6 to see what happens. Let's try that:f((x + 6) + 6) + f((x + 6) - 6) = f(x + 6)Simplifying that, we get:f(x + 12) + f(x) = f(x + 6)Hmm, so now I have another equation:f(x + 12) + f(x) = f(x + 6)Now, let's see if I can combine this with the original equation. The original equation was f(x + 6) + f(x - 6) = f(x). Maybe I can substitute f(x + 6) from the second equation into the original one.From the second equation, f(x + 6) = f(x + 12) + f(x). Let's plug that into the original equation:[f(x + 12) + f(x)] + f(x - 6) = f(x)Simplify that:f(x + 12) + f(x) + f(x - 6) = f(x)Subtract f(x) from both sides:f(x + 12) + f(x - 6) = 0Okay, so now I have:f(x + 12) + f(x - 6) = 0That's interesting. Maybe I can use this to find another relation. Let's substitute x with x + 6 in this new equation:f((x + 6) + 12) + f((x + 6) - 6) = 0Simplify:f(x + 18) + f(x) = 0So, f(x + 18) = -f(x)Alright, so f(x + 18) is equal to negative f(x). That's a useful relation. Now, let's see what happens if we shift x by another 18. Let's substitute x with x + 18 in the equation f(x + 18) = -f(x):f((x + 18) + 18) = -f(x + 18)Simplify:f(x + 36) = -f(x + 18)But from earlier, we know that f(x + 18) = -f(x). So substitute that in:f(x + 36) = -(-f(x)) = f(x)So, f(x + 36) = f(x). That means the function f is periodic with period 36. So, 36 is a period of the function.But the question is asking for the least common positive period. So, is 36 the smallest possible period, or could there be a smaller period?Let me think. If 36 is a period, then any divisor of 36 could potentially be a period as well. The options given include 12, 18, 24, 36, and 72. So, 12, 18, 24, and 36 are all divisors of 36. So, could one of these smaller numbers be the period?Let me check. If f(x + 18) = -f(x), then f(x + 36) = f(x). So, if I consider f(x + 18) = -f(x), then f(x + 36) = f(x). So, is 18 a period? Let's see.If 18 were a period, then f(x + 18) = f(x). But we have f(x + 18) = -f(x). So, unless f(x) is zero, which is trivial, f(x + 18) cannot be equal to f(x) because it's equal to -f(x). So, unless f(x) is zero, 18 cannot be a period. Therefore, 18 is not a period.Similarly, let's check 12. Suppose 12 is a period. Then f(x + 12) = f(x). But from earlier, we have f(x + 12) + f(x - 6) = 0. If f(x + 12) = f(x), then f(x) + f(x - 6) = 0. So, f(x) = -f(x - 6). Then, shifting x by 6, we get f(x + 6) = -f(x). But from the original equation, f(x + 6) + f(x - 6) = f(x). If f(x + 6) = -f(x), then substituting, we have -f(x) + f(x - 6) = f(x). So, f(x - 6) = 2f(x). Hmm, that seems problematic because if f(x - 6) = 2f(x), then f(x) = 2f(x + 6). But from earlier, f(x + 6) = -f(x). So, f(x) = 2*(-f(x)) = -2f(x). So, f(x) = -2f(x) implies 3f(x) = 0, so f(x) = 0. So, again, only the zero function would satisfy this, which is trivial. Therefore, 12 cannot be a period unless f is zero.Similarly, let's check 24. Suppose 24 is a period. Then f(x + 24) = f(x). Let's see if that can hold. From f(x + 18) = -f(x), if we shift x by 6, we get f(x + 24) = -f(x + 6). But if 24 is a period, then f(x + 24) = f(x). So, f(x) = -f(x + 6). So, f(x + 6) = -f(x). Then, from the original equation, f(x + 6) + f(x - 6) = f(x). Substituting f(x + 6) = -f(x), we get -f(x) + f(x - 6) = f(x). So, f(x - 6) = 2f(x). Then, shifting x by 6, f(x) = 2f(x + 6). But f(x + 6) = -f(x), so f(x) = 2*(-f(x)) = -2f(x). Hence, 3f(x) = 0, so f(x) = 0. Again, only the zero function, which is trivial. So, 24 cannot be a period unless f is zero.Therefore, the smallest period that works for all such functions is 36. So, the answer should be 36, which is option D.Wait, just to make sure, let me think again. We have f(x + 36) = f(x), so 36 is definitely a period. We checked smaller periods, and they only work if the function is zero, which is trivial. So, for non-trivial functions, 36 is the minimal period.Yeah, I think that's solid. So, the least common positive period is 36.**Final Answer**boxed{D}

answer：Okay, so I have this problem about a parabola and an ellipse. Let me try to understand what it's asking. First, the parabola is given by ( C_1: y^2 = 8x ). I remember that the standard form of a parabola that opens to the right is ( y^2 = 4px ), where ( p ) is the distance from the vertex to the focus. Comparing this to the given equation, ( 4p = 8 ), so ( p = 2 ). That means the focus ( F ) of the parabola is at ( (2, 0) ). Got that.Next, there's an ellipse ( C_2: frac{x^2}{m^2} + frac{y^2}{n^2} = 1 ) where ( m > n > 0 ). One of its foci coincides with the focus of the parabola, so that focus is also at ( (2, 0) ). For an ellipse, the distance from the center to each focus is ( c ), where ( c^2 = m^2 - n^2 ). Since one focus is at ( (2, 0) ), the center of the ellipse must be somewhere along the x-axis. But wait, the standard form of an ellipse is centered at the origin unless specified otherwise. Hmm, actually, the problem doesn't mention the center, so I think it's safe to assume the ellipse is also centered at the origin because the focus is at ( (2, 0) ). So, the center is at ( (0, 0) ), and the foci are at ( (pm c, 0) ). Since one focus is at ( (2, 0) ), then ( c = 2 ). Therefore, ( c^2 = 4 = m^2 - n^2 ). So, ( m^2 - n^2 = 4 ). I'll note that down as equation (1).Now, the problem states that there exist two distinct points on the ellipse ( C_2 ) that are symmetric with respect to the line ( l: y = frac{1}{4}x + frac{1}{3} ). I need to find the range of the eccentricity ( e ) of the ellipse ( C_2 ).Eccentricity ( e ) of an ellipse is given by ( e = frac{c}{m} ). Since ( c = 2 ), ( e = frac{2}{m} ). So, if I can find the range of ( m ), I can find the range of ( e ).To find the range of ( m ), I need to analyze the condition that there are two distinct points on the ellipse symmetric with respect to the line ( l ). Let me think about what that means.If two points are symmetric with respect to a line, then that line is the perpendicular bisector of the segment joining those two points. So, if I have two points ( M(x_1, y_1) ) and ( N(x_2, y_2) ) on the ellipse, then the midpoint of ( MN ) lies on the line ( l ), and the line ( MN ) is perpendicular to ( l ).The slope of line ( l ) is ( frac{1}{4} ), so the slope of the perpendicular line ( MN ) is ( -4 ). Therefore, the equation of line ( MN ) is ( y = -4x + lambda ) for some constant ( lambda ).Now, I can set up the system of equations consisting of the ellipse and the line ( MN ):[begin{cases}frac{x^2}{m^2} + frac{y^2}{n^2} = 1 y = -4x + lambdaend{cases}]Substituting ( y = -4x + lambda ) into the ellipse equation:[frac{x^2}{m^2} + frac{(-4x + lambda)^2}{n^2} = 1]Expanding this:[frac{x^2}{m^2} + frac{16x^2 - 8lambda x + lambda^2}{n^2} = 1]Combine like terms:[left( frac{1}{m^2} + frac{16}{n^2} right)x^2 - frac{8lambda}{n^2}x + frac{lambda^2}{n^2} - 1 = 0]This is a quadratic equation in ( x ). For there to be two distinct points, the discriminant must be positive. The discriminant ( D ) of ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ). So, let's compute the discriminant:Let me denote:( a = frac{1}{m^2} + frac{16}{n^2} )( b = -frac{8lambda}{n^2} )( c = frac{lambda^2}{n^2} - 1 )So,[D = left( -frac{8lambda}{n^2} right)^2 - 4 left( frac{1}{m^2} + frac{16}{n^2} right) left( frac{lambda^2}{n^2} - 1 right)]Simplify:[D = frac{64lambda^2}{n^4} - 4 left( frac{1}{m^2} + frac{16}{n^2} right) left( frac{lambda^2}{n^2} - 1 right)]Expanding the second term:[4 left( frac{1}{m^2} + frac{16}{n^2} right) left( frac{lambda^2}{n^2} - 1 right) = 4 left( frac{lambda^2}{m^2 n^2} - frac{1}{m^2} + frac{16lambda^2}{n^4} - frac{16}{n^2} right )]So,[D = frac{64lambda^2}{n^4} - 4 left( frac{lambda^2}{m^2 n^2} - frac{1}{m^2} + frac{16lambda^2}{n^4} - frac{16}{n^2} right )]Simplify term by term:First term: ( frac{64lambda^2}{n^4} )Subtracting the second term:- ( 4 cdot frac{lambda^2}{m^2 n^2} = frac{4lambda^2}{m^2 n^2} )- ( 4 cdot left( -frac{1}{m^2} right ) = -frac{4}{m^2} )- ( 4 cdot frac{16lambda^2}{n^4} = frac{64lambda^2}{n^4} )- ( 4 cdot left( -frac{16}{n^2} right ) = -frac{64}{n^2} )Putting it all together:[D = frac{64lambda^2}{n^4} - frac{4lambda^2}{m^2 n^2} + frac{4}{m^2} - frac{64lambda^2}{n^4} + frac{64}{n^2}]Simplify:The ( frac{64lambda^2}{n^4} ) and ( -frac{64lambda^2}{n^4} ) cancel out.So,[D = -frac{4lambda^2}{m^2 n^2} + frac{4}{m^2} + frac{64}{n^2}]Factor out 4:[D = 4 left( -frac{lambda^2}{m^2 n^2} + frac{1}{m^2} + frac{16}{n^2} right )]For the quadratic to have two distinct real roots, ( D > 0 ). So,[-frac{lambda^2}{m^2 n^2} + frac{1}{m^2} + frac{16}{n^2} > 0]Multiply both sides by ( m^2 n^2 ) (which is positive, so inequality sign remains):[-lambda^2 + n^2 + 16 m^2 > 0]So,[16 m^2 + n^2 - lambda^2 > 0 quad (2)]Okay, that's one condition.Now, the midpoint of ( MN ) lies on the line ( l ). Let's find the midpoint.From the quadratic equation ( ax^2 + bx + c = 0 ), the sum of roots ( x_1 + x_2 = -b/a ). So,[x_1 + x_2 = frac{8 lambda}{n^2} / left( frac{1}{m^2} + frac{16}{n^2} right ) = frac{8 lambda}{n^2} cdot frac{m^2 n^2}{n^2 + 16 m^2} = frac{8 lambda m^2}{n^2 + 16 m^2}]Similarly, the sum of ( y_1 + y_2 ) can be found since ( y = -4x + lambda ):[y_1 + y_2 = -4(x_1 + x_2) + 2lambda = -4 cdot frac{8 lambda m^2}{n^2 + 16 m^2} + 2lambda = -frac{32 lambda m^2}{n^2 + 16 m^2} + 2lambda]Factor out ( lambda ):[y_1 + y_2 = lambda left( -frac{32 m^2}{n^2 + 16 m^2} + 2 right ) = lambda left( frac{-32 m^2 + 2(n^2 + 16 m^2)}{n^2 + 16 m^2} right ) = lambda left( frac{-32 m^2 + 2 n^2 + 32 m^2}{n^2 + 16 m^2} right ) = frac{2 lambda n^2}{n^2 + 16 m^2}]So, the midpoint ( (h, k) ) is:[h = frac{x_1 + x_2}{2} = frac{4 lambda m^2}{n^2 + 16 m^2}][k = frac{y_1 + y_2}{2} = frac{lambda n^2}{n^2 + 16 m^2}]Since this midpoint lies on line ( l: y = frac{1}{4}x + frac{1}{3} ), substituting ( h ) and ( k ):[frac{lambda n^2}{n^2 + 16 m^2} = frac{1}{4} cdot frac{4 lambda m^2}{n^2 + 16 m^2} + frac{1}{3}]Simplify the right-hand side:[frac{lambda n^2}{n^2 + 16 m^2} = frac{lambda m^2}{n^2 + 16 m^2} + frac{1}{3}]Subtract ( frac{lambda m^2}{n^2 + 16 m^2} ) from both sides:[frac{lambda n^2 - lambda m^2}{n^2 + 16 m^2} = frac{1}{3}]Factor ( lambda ):[frac{lambda (n^2 - m^2)}{n^2 + 16 m^2} = frac{1}{3}]From equation (1), ( m^2 - n^2 = 4 ), so ( n^2 - m^2 = -4 ). Substitute:[frac{lambda (-4)}{n^2 + 16 m^2} = frac{1}{3}]Multiply both sides by ( n^2 + 16 m^2 ):[-4 lambda = frac{n^2 + 16 m^2}{3}]So,[lambda = -frac{n^2 + 16 m^2}{12}]Now, substitute ( lambda ) back into condition (2):[16 m^2 + n^2 - lambda^2 > 0]First, compute ( lambda^2 ):[lambda^2 = left( -frac{n^2 + 16 m^2}{12} right )^2 = frac{(n^2 + 16 m^2)^2}{144}]So,[16 m^2 + n^2 - frac{(n^2 + 16 m^2)^2}{144} > 0]Multiply both sides by 144 to eliminate the denominator:[144(16 m^2 + n^2) - (n^2 + 16 m^2)^2 > 0]Let me denote ( A = n^2 + 16 m^2 ). Then,[144(16 m^2 + n^2) - A^2 > 0]But ( A = n^2 + 16 m^2 ), so:[144 A - A^2 > 0]Factor:[A(144 - A) > 0]So, this inequality holds when ( 0 < A < 144 ).But ( A = n^2 + 16 m^2 ). Since ( n^2 > 0 ) and ( m^2 > 0 ), ( A > 0 ). So, the condition is ( A < 144 ):[n^2 + 16 m^2 < 144]But from equation (1), ( n^2 = m^2 - 4 ). Substitute:[(m^2 - 4) + 16 m^2 < 144]Simplify:[17 m^2 - 4 < 144][17 m^2 < 148][m^2 < frac{148}{17}][m^2 < frac{148}{17} approx 8.7059][m < sqrt{frac{148}{17}} = frac{sqrt{148}}{sqrt{17}} = frac{2sqrt{37}}{sqrt{17}} = frac{2sqrt{629}}{17}]Wait, hold on. Let me compute ( sqrt{148} ). ( 148 = 4 times 37 ), so ( sqrt{148} = 2sqrt{37} ). Therefore,[m < frac{2sqrt{37}}{sqrt{17}} = frac{2sqrt{37 times 17}}{17} = frac{2sqrt{629}}{17}]So, ( m < frac{2sqrt{629}}{17} ). But we also know from equation (1) that ( m^2 = n^2 + 4 ). Since ( n > 0 ), ( m > sqrt{4} = 2 ). So, ( m > 2 ).Therefore, combining both inequalities:[2 < m < frac{2sqrt{629}}{17}]Now, the eccentricity ( e = frac{c}{m} = frac{2}{m} ). So, as ( m ) increases, ( e ) decreases, and as ( m ) decreases, ( e ) increases.So, when ( m ) is at its minimum value ( 2 ), ( e ) is maximum ( frac{2}{2} = 1 ). But wait, an ellipse can't have eccentricity equal to 1, because that's a parabola. So, actually, the maximum value of ( e ) is less than 1.Wait, hold on. The eccentricity of an ellipse is always less than 1. So, ( e < 1 ). So, when ( m ) approaches 2 from above, ( e ) approaches 1 from below. When ( m ) is at its maximum value ( frac{2sqrt{629}}{17} ), ( e ) is at its minimum value:[e_{text{min}} = frac{2}{frac{2sqrt{629}}{17}} = frac{17}{sqrt{629}} = frac{sqrt{629}}{37}]Because ( frac{17}{sqrt{629}} = frac{17 sqrt{629}}{629} = frac{sqrt{629}}{37} ).So, the range of ( e ) is:[frac{sqrt{629}}{37} < e < 1]Let me double-check my steps to make sure I didn't make a mistake.1. Found the focus of the parabola correctly: ( (2, 0) ).2. Set up the ellipse with ( c = 2 ), so ( m^2 - n^2 = 4 ).3. Considered the line of symmetry, found the equation of the perpendicular line, substituted into the ellipse, found the discriminant condition.4. Calculated the midpoint and substituted into the line equation to find ( lambda ).5. Plugged ( lambda ) back into the discriminant condition, leading to an inequality involving ( m ).6. Solved for ( m ) and then found the corresponding eccentricity range.Everything seems to check out. So, the range of the eccentricity ( e ) is ( frac{sqrt{629}}{37} < e < 1 ).**Final Answer**The range of the eccentricity ( e ) of the ellipse ( C_{2} ) is boxed{dfrac{sqrt{629}}{37} < e < 1}.

answer：First, I need to determine how many marbles Carolyn had initially. She ended up with 5 marbles after sharing 42 marbles with Diana. To find the initial number of marbles, I add the marbles she gave away to the marbles she has left.So, the initial number of marbles is 42 plus 5, which equals 47.

answer：Okay, so I have this problem about an ellipse with foci at (-1, 0) and (1, 0). The equation of the ellipse is given as x²/a² + y²/b² = 1, where a > b > 0. There's a point P on the ellipse such that PF₂ is perpendicular to F₁F₂, and the difference in distances from P to F₁ and F₂ is a/2. First, I need to find the equation of the ellipse. Let me recall some properties of ellipses. The distance between the two foci is 2c, so here, since the foci are at (-1, 0) and (1, 0), the distance between them is 2, so c = 1. For an ellipse, we know that a² = b² + c². So once I find a and b, I can write the equation. The problem says that PF₂ is perpendicular to F₁F₂. Since F₁F₂ is the line segment from (-1, 0) to (1, 0), which is along the x-axis, PF₂ being perpendicular means that PF₂ is vertical. So point P must lie somewhere along the vertical line passing through F₂, which is x = 1. So, point P is (1, y) for some y. Since P is on the ellipse, plugging into the ellipse equation: (1)²/a² + y²/b² = 1, so y² = b²(1 - 1/a²). Now, the distances |PF₁| and |PF₂|. Since F₁ is (-1, 0) and F₂ is (1, 0), and P is (1, y), let's compute these distances. |PF₂| is the distance from (1, y) to (1, 0), which is just |y|. |PF₁| is the distance from (1, y) to (-1, 0). Using the distance formula: sqrt[(1 - (-1))² + (y - 0)²] = sqrt[(2)² + y²] = sqrt(4 + y²). The problem states that |PF₁| - |PF₂| = a/2. So, sqrt(4 + y²) - |y| = a/2. But since y² is positive, |y| is just the square root of y², so we can write sqrt(4 + y²) - sqrt(y²) = a/2. Let me denote sqrt(y²) as |y|, but since y can be positive or negative, but in the context of the ellipse, it's symmetric, so we can assume y is positive without loss of generality. So, sqrt(4 + y²) - y = a/2. Let me solve for y. Let me set sqrt(4 + y²) = y + a/2. Then, squaring both sides: 4 + y² = y² + a y + a²/4. Simplify: 4 = a y + a²/4. So, a y = 4 - a²/4. Therefore, y = (4 - a²/4)/a = (16 - a²)/4a. But earlier, we had y² = b²(1 - 1/a²). Let's compute y² from the expression above. y = (16 - a²)/(4a), so y² = [(16 - a²)/(4a)]² = (256 - 32a² + a⁴)/(16a²). On the other hand, y² = b²(1 - 1/a²) = b²(a² - 1)/a². So, equate the two expressions for y²: (256 - 32a² + a⁴)/(16a²) = b²(a² - 1)/a². Multiply both sides by 16a²: 256 - 32a² + a⁴ = 16b²(a² - 1). But we know from the ellipse that a² = b² + c², and c = 1, so a² = b² + 1. Therefore, b² = a² - 1. Substitute b² into the equation: 256 - 32a² + a⁴ = 16(a² - 1)(a² - 1). Simplify the right-hand side: 16(a² - 1)² = 16(a⁴ - 2a² + 1). So, the equation becomes: 256 - 32a² + a⁴ = 16a⁴ - 32a² + 16. Bring all terms to the left-hand side: 256 - 32a² + a⁴ - 16a⁴ + 32a² - 16 = 0. Simplify: (256 - 16) + (-32a² + 32a²) + (a⁴ - 16a⁴) = 0 240 - 15a⁴ = 0 So, 15a⁴ = 240 a⁴ = 16 Therefore, a² = 4 (since a > 0). So, a = 2. Then, since a² = b² + 1, b² = 4 - 1 = 3. Therefore, the equation of the ellipse is x²/4 + y²/3 = 1. Okay, so that's part (1). Now, part (2): Point B is the top vertex of the ellipse, which is (0, b) = (0, sqrt(3)). A line l passes through F₂(1, 0) and intersects the ellipse at two distinct points M and N. We need to determine if there exists such a line l where the ratio of the areas of triangles BF₂M and BF₂N is 2. If it exists, find the equation of line l; otherwise, explain why not. First, let me visualize this. Point B is at (0, sqrt(3)). F₂ is at (1, 0). The line l passes through F₂ and intersects the ellipse at M and N. So, M and N are points on the ellipse, and line l connects F₂ to M and N. We need the ratio of the areas of triangles BF₂M and BF₂N to be 2. Since both triangles share the same base BF₂, the ratio of their areas would be equal to the ratio of their heights from BF₂. But actually, wait, no. Let me think. Triangles BF₂M and BF₂N share the common vertex at B and have bases F₂M and F₂N. So, the area ratio would be equal to the ratio of the lengths of F₂M and F₂N. Because the height from B to the line F₂M is the same as the height from B to F₂N if the line l is such that both M and N are on the same line. Wait, no, actually, since both M and N are on line l, which passes through F₂, the triangles BF₂M and BF₂N share the same vertex B and have their bases on the same line l. Therefore, the ratio of their areas is equal to the ratio of the lengths of F₂M and F₂N. Wait, but actually, the area of a triangle is 1/2 * base * height. Here, the base is F₂M or F₂N, and the height is the distance from B to the line l. But since both triangles share the same height (distance from B to line l), the ratio of their areas is equal to the ratio of their bases, which are |F₂M| and |F₂N|. Therefore, if the ratio of areas is 2, then |F₂M| / |F₂N| = 2. Alternatively, since the line passes through F₂, the points M and N are on opposite sides of F₂, so perhaps one is on one side and the other is on the opposite side. Therefore, the distances could be considered with signs, but since we are talking about lengths, they are positive. Alternatively, perhaps the ratio is |F₂M| / |F₂N| = 2 or |F₂N| / |F₂M| = 2. So, we need to find a line l through F₂ such that one of the intersection points is twice as far from F₂ as the other. So, let me parametrize the line l. Let me assume that the line l has a slope m, so its equation is y = m(x - 1), since it passes through F₂(1, 0). Alternatively, to avoid dealing with vertical lines, I can parametrize it as x = ty + 1, where t is the parameter. But perhaps using slope-intercept is easier. Let me proceed with the slope-intercept form: y = m(x - 1). Now, substitute this into the ellipse equation x²/4 + y²/3 = 1. So, x²/4 + [m(x - 1)]² / 3 = 1. Let me expand this: x²/4 + (m²(x² - 2x + 1))/3 = 1 Multiply through by 12 to eliminate denominators: 3x² + 4m²(x² - 2x + 1) = 12 Expand: 3x² + 4m²x² - 8m²x + 4m² = 12 Combine like terms: (3 + 4m²)x² - 8m²x + (4m² - 12) = 0 This is a quadratic in x: (3 + 4m²)x² - 8m²x + (4m² - 12) = 0 Let me denote this as Ax² + Bx + C = 0, where A = 3 + 4m² B = -8m² C = 4m² - 12 The solutions to this quadratic will give the x-coordinates of points M and N. Let me denote the roots as x₁ and x₂. From Vieta's formulas: x₁ + x₂ = -B/A = (8m²)/(3 + 4m²) x₁x₂ = C/A = (4m² - 12)/(3 + 4m²) Now, the points M and N are on the line y = m(x - 1), so their coordinates are (x₁, m(x₁ - 1)) and (x₂, m(x₂ - 1)). We need to find the distances |F₂M| and |F₂N|. Since F₂ is at (1, 0), the distance from F₂ to M is sqrt[(x₁ - 1)² + (y₁ - 0)²] = sqrt[(x₁ - 1)² + (m(x₁ - 1))²] = |x₁ - 1| * sqrt(1 + m²). Similarly, |F₂N| = |x₂ - 1| * sqrt(1 + m²). Since sqrt(1 + m²) is a common factor, the ratio |F₂M| / |F₂N| = |x₁ - 1| / |x₂ - 1|. We need this ratio to be 2 or 1/2. But let's think about the roots x₁ and x₂. Since the line passes through F₂(1, 0), which is a focus, but not necessarily a point on the ellipse. Wait, actually, the ellipse is x²/4 + y²/3 = 1, so plugging in (1, 0): 1/4 + 0 = 1/4 ≠ 1, so F₂ is inside the ellipse, not on it. Therefore, the line passes through F₂ and intersects the ellipse at two points M and N. Therefore, the distances |F₂M| and |F₂N| are both positive, and we need their ratio to be 2. So, |x₁ - 1| / |x₂ - 1| = 2. But since the quadratic equation is in x, and the line passes through (1, 0), which is not a root (since plugging x=1 into the quadratic: (3 + 4m²)(1) - 8m²(1) + (4m² - 12) = 3 + 4m² - 8m² + 4m² - 12 = (3 - 12) + (4m² - 8m² + 4m²) = -9 + 0 = -9 ≠ 0). So, x=1 is not a root, which makes sense because F₂ is inside the ellipse. Therefore, the roots x₁ and x₂ are on either side of x=1? Not necessarily, because the line could intersect the ellipse on the same side of F₂. But since the ellipse is symmetric, it's possible that one point is on one side and the other is on the opposite side. But regardless, the distances |x₁ - 1| and |x₂ - 1| are positive. So, we have |x₁ - 1| = 2 |x₂ - 1|. But since x₁ and x₂ are roots of the quadratic, we can relate them using Vieta's formulas. Let me denote u = x - 1. Then, x = u + 1. Substitute into the quadratic equation: (3 + 4m²)(u + 1)² - 8m²(u + 1) + (4m² - 12) = 0 Expand (u + 1)²: u² + 2u + 1 So, (3 + 4m²)(u² + 2u + 1) - 8m²(u + 1) + (4m² - 12) = 0 Expand term by term: (3 + 4m²)u² + 2(3 + 4m²)u + (3 + 4m²) - 8m²u - 8m² + 4m² - 12 = 0 Combine like terms: (3 + 4m²)u² + [2(3 + 4m²) - 8m²]u + [3 + 4m² - 8m² + 4m² - 12] = 0 Simplify each bracket: For the u² term: remains (3 + 4m²)u² For the u term: 6 + 8m² - 8m² = 6 For the constant term: 3 + 4m² - 8m² + 4m² - 12 = (3 - 12) + (4m² - 8m² + 4m²) = -9 + 0 = -9 So, the equation becomes: (3 + 4m²)u² + 6u - 9 = 0 Divide through by 3: ( (3 + 4m²)/3 )u² + 2u - 3 = 0 But maybe it's better to keep it as is: (3 + 4m²)u² + 6u - 9 = 0 Let me denote this as A u² + B u + C = 0, where A = 3 + 4m² B = 6 C = -9 The roots are u₁ and u₂, which correspond to x₁ - 1 and x₂ - 1. From Vieta's formulas: u₁ + u₂ = -B/A = -6/(3 + 4m²) u₁ u₂ = C/A = -9/(3 + 4m²) We have |u₁| = 2 |u₂|. Since u₁ and u₂ are real numbers, their product is negative (since C/A is negative). Therefore, one of u₁ or u₂ is positive and the other is negative. Without loss of generality, let's assume u₁ is positive and u₂ is negative. Then, |u₁| = u₁, |u₂| = -u₂. Given that |u₁| = 2 |u₂|, so u₁ = 2 (-u₂) = -2 u₂. So, u₁ = -2 u₂. From Vieta's formulas: u₁ + u₂ = -6/(3 + 4m²) But u₁ = -2 u₂, so substituting: -2 u₂ + u₂ = -6/(3 + 4m²) - u₂ = -6/(3 + 4m²) Therefore, u₂ = 6/(3 + 4m²) Then, u₁ = -2 u₂ = -12/(3 + 4m²) Now, from the product of roots: u₁ u₂ = (-12/(3 + 4m²)) * (6/(3 + 4m²)) = (-72)/(3 + 4m²)² But from Vieta, u₁ u₂ = -9/(3 + 4m²). Therefore, (-72)/(3 + 4m²)² = -9/(3 + 4m²) Multiply both sides by (3 + 4m²)²: -72 = -9(3 + 4m²) Divide both sides by -9: 8 = 3 + 4m² Therefore, 4m² = 5 So, m² = 5/4 Thus, m = ±√(5)/2 Therefore, the slopes of line l are m = √5/2 and m = -√5/2. Therefore, the equations of line l are y = (√5/2)(x - 1) and y = (-√5/2)(x - 1). So, such lines exist, and their equations are y = ±(√5/2)(x - 1). Alternatively, to write them in standard form: For m = √5/2: y = (√5/2)x - √5/2 Multiply both sides by 2: 2y = √5 x - √5 Rearranged: √5 x - 2y - √5 = 0 Similarly, for m = -√5/2: y = (-√5/2)x + √5/2 Multiply both sides by 2: 2y = -√5 x + √5 Rearranged: √5 x + 2y - √5 = 0 But the problem asks for the equation of line l, so either form is acceptable, but perhaps the slope-intercept form is more straightforward. Therefore, the equations are y = ±(√5/2)(x - 1). So, to recap, such lines do exist, and their equations are y = ±(√5/2)(x - 1). I think that's the solution. Let me just check if I made any mistakes. Wait, when I substituted u = x - 1, I transformed the quadratic correctly. Then, using Vieta's, I set up the ratio |u₁| = 2 |u₂|, considering the signs. Then, solved for u₁ and u₂ in terms of m, and found m² = 5/4. That seems correct. Yes, I think this is correct.