answer：Okay, so I've got this system of equations to solve:[3x + 4y = 10][12x - 8y = 8]Hmm, I need to find the values of x and y that satisfy both equations. Let me think about the best way to approach this. I remember there are a few methods to solve systems of equations, like substitution, elimination, or maybe even graphing. Since these are linear equations, elimination might be a good method here because the coefficients look like they can be easily manipulated to eliminate one of the variables.Looking at the first equation, 3x + 4y = 10, and the second equation, 12x - 8y = 8, I notice that the coefficients of y in both equations are 4 and -8. If I can make them the same or opposites, I can eliminate y by adding or subtracting the equations. Let me try to make the coefficients of y opposites so that when I add the equations, y will cancel out.To do that, I can multiply the entire first equation by 2. Let's see:Multiplying the first equation by 2:[2*(3x + 4y) = 2*10]Which gives:[6x + 8y = 20]Now, the second equation is:[12x - 8y = 8]If I add these two equations together, the y terms should cancel out:[(6x + 8y) + (12x - 8y) = 20 + 8]Simplifying the left side:6x + 12x + 8y - 8y = 18xAnd the right side:20 + 8 = 28So, I have:[18x = 28]Now, solving for x:[x = frac{28}{18}]Simplify the fraction by dividing numerator and denominator by 2:[x = frac{14}{9}]Okay, so x is 14/9. Now, I need to find y. I can plug this value of x back into one of the original equations. Let me choose the first equation because the numbers seem smaller and easier to work with.First equation:[3x + 4y = 10]Substitute x = 14/9:[3*(14/9) + 4y = 10]Calculate 3*(14/9):[42/9 + 4y = 10]Simplify 42/9 to 14/3:[14/3 + 4y = 10]Now, subtract 14/3 from both sides to solve for 4y:[4y = 10 - 14/3]Convert 10 to thirds to subtract:10 = 30/3, so:[4y = 30/3 - 14/3 = 16/3]Now, divide both sides by 4 to solve for y:[y = (16/3)/4 = 16/12 = 4/3]So, y is 4/3. Let me check if these values satisfy both equations to make sure I didn't make a mistake.First equation:[3x + 4y = 10]Substitute x = 14/9 and y = 4/3:[3*(14/9) + 4*(4/3) = 42/9 + 16/3]Simplify:42/9 = 14/3, so:14/3 + 16/3 = 30/3 = 10Which matches the first equation.Second equation:[12x - 8y = 8]Substitute x = 14/9 and y = 4/3:[12*(14/9) - 8*(4/3) = 168/9 - 32/3]Simplify:168/9 = 56/3, and 32/3 is already simplified:56/3 - 32/3 = 24/3 = 8Which matches the second equation.Great, both equations are satisfied with x = 14/9 and y = 4/3. So, the solution is x = 14/9 and y = 4/3.Looking back at the options provided:A. x=14/9, y=4/3B. x=1, y=1C. x=2, y=2D. There is no solutionE. There are an infinite number of solutionsSo, option A is the correct answer.

answer：Alright, so I have this problem with four propositions about angles and lines, and I need to figure out how many of them are true. Let's take them one by one and think through each carefully.**Proposition ①: Vertically opposite angles are equal.**Okay, vertically opposite angles. I remember these are the angles opposite each other when two lines intersect. Like, if two lines cross, they form an 'X' shape, and the angles directly across from each other are vertically opposite. My teacher said they are always equal. Let me visualize it: if one angle is, say, 40 degrees, the one across should also be 40 degrees. Yeah, that makes sense because the lines are straight, so the angles have to balance out. I think this one is true.**Proposition ②: In complementary angles, one must be acute and the other obtuse.**Hmm, complementary angles add up to 90 degrees. Acute angles are less than 90 degrees, and obtuse angles are more than 90 degrees. Wait, if two angles add up to 90 degrees, can one of them be obtuse? If one angle is more than 90 degrees, the other would have to be negative to make the sum 90, which doesn't make sense. So both angles must be less than 90 degrees, meaning both are acute. So this proposition is saying one has to be obtuse, but that's not true. Both are acute. So this one is false.**Proposition ③: Corresponding angles are equal, then the two lines are parallel.**Corresponding angles... Okay, when a transversal crosses two lines, corresponding angles are in the same position relative to the lines and the transversal. If those angles are equal, does that mean the lines are parallel? I think that's one of the theorems. If corresponding angles are equal, then the lines must be parallel. Yeah, that sounds right. So this one is true.Wait, hold on. The proposition says "corresponding angles are equal, then the two lines are parallel." But isn't that only true if the lines are cut by a transversal? The proposition doesn't mention a transversal. Maybe that's important. Hmm, I'm not sure. Maybe it's implied because we're talking about corresponding angles. I think in geometry, corresponding angles are defined in the context of a transversal. So maybe it's okay. I'll go with true for now.**Proposition ④: The bisectors of the corresponding angles are perpendicular to each other.**Bisectors of corresponding angles... So if you have two corresponding angles, and you bisect each of them, are those bisectors perpendicular? Let me think. If the lines are parallel, then the corresponding angles are equal, and their bisectors would create smaller angles. Would those bisectors be perpendicular? I don't think so. If the original angles are, say, 60 degrees each, their bisectors would create 30-degree angles. So the angle between the bisectors would be 30 degrees, not 90. So they wouldn't be perpendicular. So this proposition is false.Wait, but what if the lines aren't parallel? If the lines aren't parallel, then the corresponding angles aren't equal, and their bisectors might not have any particular relationship. So in general, this doesn't hold. So yeah, I think this one is false.Okay, so summarizing:① True② False③ True④ FalseSo out of four propositions, two are true. That would be option B: 2.But wait, earlier I was a bit unsure about proposition ③ because it didn't mention a transversal. Let me double-check that. Corresponding angles are defined when a transversal crosses two lines. So if corresponding angles are equal, then by the converse of the corresponding angles postulate, the lines are parallel. So yeah, I think it's correct. So ③ is true.So yeah, two true propositions: ① and ③.

answer：Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. It says that the geometric sequence {a_n} is increasing, the first term a_1 is -2, and there's this equation: 3(a_n + a_{n+2}) = 10a_{n+1}. I need to find the common ratio q, and it's given that the answer is 1/3. But I should figure out why that's the case.First, let's recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, q. So, in general, a_n = a_1 * q^{n-1}. Since a_1 is -2, the nth term is -2 * q^{n-1}.Now, the sequence is increasing. Hmm, that's interesting because the first term is negative. So, if it's increasing, each subsequent term must be getting larger, meaning less negative or becoming positive. But since it's a geometric sequence, the behavior depends on the common ratio q.If q is greater than 1, the terms would become more negative if a_1 is negative, right? Because multiplying a negative number by a number greater than 1 makes it more negative. So, that would make the sequence decreasing, not increasing. On the other hand, if q is between 0 and 1, multiplying a negative number by a fraction between 0 and 1 makes it less negative, so the terms increase. So, that tells me that 0 < q < 1.Okay, so that's a clue. Now, let's look at the equation given: 3(a_n + a_{n+2}) = 10a_{n+1}. I need to translate this into something involving q.Since it's a geometric sequence, a_{n} = a_1 * q^{n-1}, a_{n+1} = a_1 * q^{n}, and a_{n+2} = a_1 * q^{n+1}. So, let's substitute these into the equation.3(a_n + a_{n+2}) = 10a_{n+1}Substituting the expressions:3(a_1 * q^{n-1} + a_1 * q^{n+1}) = 10 * a_1 * q^{n}I can factor out a_1 * q^{n-1} from the left side:3(a_1 * q^{n-1} (1 + q^2)) = 10 * a_1 * q^{n}Let me write that out:3 * a_1 * q^{n-1} * (1 + q^2) = 10 * a_1 * q^{n}Now, I can divide both sides by a_1 * q^{n-1}, assuming a_1 ≠ 0 and q ≠ 0, which they aren't in this case because a_1 is -2 and q is between 0 and 1.So, dividing both sides:3 * (1 + q^2) = 10 * qThat simplifies to:3 + 3q^2 = 10qLet me rearrange this into a standard quadratic equation:3q^2 - 10q + 3 = 0Okay, so now I have a quadratic equation in terms of q. Let's solve for q.Quadratic equation: 3q^2 - 10q + 3 = 0We can use the quadratic formula: q = [10 ± sqrt(100 - 36)] / 6Calculating the discriminant:sqrt(100 - 36) = sqrt(64) = 8So, q = [10 ± 8] / 6That gives two solutions:q = (10 + 8)/6 = 18/6 = 3q = (10 - 8)/6 = 2/6 = 1/3So, q can be 3 or 1/3.But earlier, I concluded that since the sequence is increasing and a_1 is negative, q must be between 0 and 1. So, q = 3 is greater than 1, which would make the sequence decreasing because each term would be more negative. But the problem says it's increasing, so q must be 1/3.Let me verify that. If q is 1/3, then each term is multiplied by 1/3, so starting from -2:a_1 = -2a_2 = -2 * (1/3) = -2/3a_3 = -2/3 * (1/3) = -2/9a_4 = -2/9 * (1/3) = -2/27And so on. So, each term is less negative than the previous one, meaning the sequence is increasing. That makes sense.If q were 3, then:a_1 = -2a_2 = -2 * 3 = -6a_3 = -6 * 3 = -18a_4 = -18 * 3 = -54This is clearly decreasing because each term is more negative. So, that's not what we want.Therefore, the common ratio q must be 1/3.I think that's solid. Let me recap:1. Recognize it's a geometric sequence with a negative first term and it's increasing, so 0 < q < 1.2. Substitute the terms into the given equation and simplify to get a quadratic equation.3. Solve the quadratic equation to find possible q values.4. Discard the q value that doesn't fit the increasing condition.So, yeah, q = 1/3 is the answer.**Final Answer**The common ratio ( q ) is boxed{dfrac{1}{3}}.

answer：Okay, so I have this problem about an ellipse E centered at the origin with its axes of symmetry along the x-axis and y-axis. It passes through two points, A(0, -2) and B(3/2, -1). I need to find the equation of the ellipse E, and then prove that a certain line passes through a fixed point. Let me try to tackle this step by step.Starting with part (1): Finding the equation of ellipse E.I remember that the general equation of an ellipse centered at the origin with major and minor axes along the coordinate axes is (x²/a²) + (y²/b²) = 1, where a and b are the semi-major and semi-minor axes. Since the ellipse passes through points A and B, I can plug these points into the equation to find a and b.Point A is (0, -2). Plugging into the ellipse equation:(0²/a²) + ((-2)²/b²) = 10 + (4/b²) = 1So, 4/b² = 1 => b² = 4 => b = 2.Okay, so the semi-minor axis is 2. Now, let's use point B(3/2, -1). Plugging into the ellipse equation:((3/2)²/a²) + ((-1)²/b²) = 1(9/4)/a² + 1/4 = 1We already know b² = 4, so 1/b² = 1/4. So, substituting:(9/4)/a² + 1/4 = 1Let me subtract 1/4 from both sides:(9/4)/a² = 3/4Multiply both sides by a²:9/4 = (3/4)a²Multiply both sides by 4:9 = 3a²Divide both sides by 3:a² = 3 => a = sqrt(3)So, the equation of the ellipse is (x²/3) + (y²/4) = 1.Wait, let me double-check. If a² is 3, then the semi-major axis is sqrt(3), and b² is 4, so semi-minor axis is 2. Since 4 > 3, the major axis is along the y-axis, which makes sense because point A is on the y-axis.Okay, that seems correct. So, part (1) is done.Moving on to part (2): This seems more involved. Let me parse the problem.We have a line passing through point P(1, -2) intersecting the ellipse E at points M and N. Then, a line passing through M parallel to the x-axis intersects the line segment AB at point T. Point H satisfies vector MT = vector TH, meaning H is such that T is the midpoint of MH. We need to prove that the line HN passes through a fixed point.Hmm. So, step by step:1. Find the equation of the line passing through P(1, -2) with some slope, intersecting the ellipse at M and N.2. For each such line, find points M and N.3. From point M, draw a horizontal line (parallel to x-axis) which intersects segment AB at T.4. Find point H such that MT = TH, so H is the reflection of M over T.5. Then, show that the line HN passes through a fixed point, regardless of the slope of the initial line through P.This seems like it might involve parametric equations, or perhaps using parametric forms of the ellipse.First, let me find the equation of line segment AB. Points A(0, -2) and B(3/2, -1). The slope of AB is ( -1 - (-2) ) / ( 3/2 - 0 ) = (1)/(3/2) = 2/3.So, the equation of AB is y - (-2) = (2/3)(x - 0), so y = (2/3)x - 2.So, any horizontal line through M (which has the same y-coordinate as M) will intersect AB at T. So, T has the same y-coordinate as M, and lies on AB.Therefore, if M is (x1, y1), then T is ( (y1 + 2) * (3/2), y1 ). Because from AB's equation, y = (2/3)x - 2, so solving for x when y = y1:y1 = (2/3)x - 2 => x = (y1 + 2) * (3/2).So, T is ( (3/2)(y1 + 2), y1 ).Then, point H is such that vector MT = vector TH. So, H is the reflection of M over T. So, if T is the midpoint between M and H, then H = 2T - M.So, if M is (x1, y1), then T is ( (3/2)(y1 + 2), y1 ), so H is ( 2*(3/2)(y1 + 2) - x1, 2*y1 - y1 ) = ( 3(y1 + 2) - x1, y1 ).So, H is ( 3y1 + 6 - x1, y1 ).Now, we need to find the equation of line HN, where N is another intersection point of the line through P with the ellipse.Wait, but N is another point on the ellipse, so if the line through P intersects the ellipse at M and N, then N is another point (x2, y2) on the ellipse.So, the line through P(1, -2) with slope k is y + 2 = k(x - 1). Let's denote this line as y = kx - k - 2.This line intersects the ellipse (x²/3) + (y²/4) = 1.Substituting y = kx - k - 2 into the ellipse equation:x²/3 + (kx - k - 2)² /4 = 1Let me expand this:x²/3 + [ (kx - k - 2)^2 ] /4 = 1First, compute (kx - k - 2)^2:= (kx)^2 + (-k - 2)^2 + 2*(kx)*(-k - 2)= k²x² + (k² + 4k + 4) + (-2k(k + 2))x= k²x² + (k² + 4k + 4) - 2k(k + 2)xSo, substituting back:x²/3 + [k²x² + (k² + 4k + 4) - 2k(k + 2)x ] /4 = 1Multiply through by 12 to eliminate denominators:4x² + 3[ k²x² + (k² + 4k + 4) - 2k(k + 2)x ] = 12Expand:4x² + 3k²x² + 3(k² + 4k + 4) - 6k(k + 2)x = 12Combine like terms:(4 + 3k²)x² - 6k(k + 2)x + 3(k² + 4k + 4) - 12 = 0Simplify the constant term:3(k² + 4k + 4) - 12 = 3k² + 12k + 12 - 12 = 3k² + 12kSo, the quadratic equation is:(4 + 3k²)x² - 6k(k + 2)x + (3k² + 12k) = 0Let me write this as:(3k² + 4)x² - 6k(k + 2)x + 3k(k + 4) = 0Wait, let me check:Wait, 3(k² + 4k + 4) is 3k² + 12k + 12, and subtracting 12 gives 3k² + 12k.So, yes, the equation is:(3k² + 4)x² - 6k(k + 2)x + 3k(k + 4) = 0Let me denote this as:A x² + B x + C = 0, whereA = 3k² + 4B = -6k(k + 2)C = 3k(k + 4)So, the solutions are x = [6k(k + 2) ± sqrt( [6k(k + 2)]² - 4*(3k² + 4)*3k(k + 4) ) ] / [2*(3k² + 4)]But perhaps instead of computing the roots, which might be messy, I can use Vieta's formula.So, for quadratic equation A x² + B x + C = 0,x1 + x2 = -B/A = [6k(k + 2)] / (3k² + 4)x1 * x2 = C/A = [3k(k + 4)] / (3k² + 4)Similarly, since y = kx - k - 2, then y1 = kx1 - k - 2, y2 = kx2 - k - 2.So, y1 + y2 = k(x1 + x2) - 2k - 4= k*[6k(k + 2)/(3k² + 4)] - 2k - 4Similarly, y1 * y2 = [kx1 - k - 2][kx2 - k - 2]= k²x1x2 - k(k + 2)(x1 + x2) + (k + 2)^2So, let me compute y1 + y2 and y1 * y2.First, y1 + y2:= k*(6k(k + 2)/(3k² + 4)) - 2k - 4= [6k²(k + 2)] / (3k² + 4) - 2k - 4Let me write 2k + 4 as [2k + 4]*(3k² + 4)/(3k² + 4) to have a common denominator.So,= [6k²(k + 2) - (2k + 4)(3k² + 4)] / (3k² + 4)Let me expand the numerator:6k²(k + 2) = 6k³ + 12k²(2k + 4)(3k² + 4) = 6k³ + 8k + 12k² + 16So, numerator:6k³ + 12k² - (6k³ + 8k + 12k² + 16) = 6k³ + 12k² - 6k³ - 8k - 12k² -16 = -8k -16Thus,y1 + y2 = (-8k -16)/(3k² + 4)Similarly, y1 * y2:= k²x1x2 - k(k + 2)(x1 + x2) + (k + 2)^2We have x1x2 = 3k(k + 4)/(3k² + 4)x1 + x2 = 6k(k + 2)/(3k² + 4)So,= k²*(3k(k + 4)/(3k² + 4)) - k(k + 2)*(6k(k + 2)/(3k² + 4)) + (k + 2)^2Simplify each term:First term: 3k³(k + 4)/(3k² + 4)Second term: -6k²(k + 2)^2/(3k² + 4)Third term: (k + 2)^2So, combining the first two terms over the same denominator:[3k³(k + 4) - 6k²(k + 2)^2]/(3k² + 4) + (k + 2)^2Let me compute the numerator:3k³(k + 4) = 3k⁴ + 12k³6k²(k + 2)^2 = 6k²(k² + 4k + 4) = 6k⁴ + 24k³ + 24k²So, numerator:3k⁴ + 12k³ - (6k⁴ + 24k³ + 24k²) = 3k⁴ +12k³ -6k⁴ -24k³ -24k² = -3k⁴ -12k³ -24k²Thus,= (-3k⁴ -12k³ -24k²)/(3k² + 4) + (k + 2)^2Factor numerator:-3k²(k² + 4k + 8)/(3k² + 4) + (k + 2)^2Hmm, not sure if that helps. Alternatively, let me factor out -3k²:= -3k²(k² + 4k + 8)/(3k² + 4) + (k + 2)^2Not sure, maybe better to leave it as is.So, y1 * y2 = (-3k⁴ -12k³ -24k²)/(3k² + 4) + (k + 2)^2Let me compute (k + 2)^2 = k² + 4k + 4So, combining:= [(-3k⁴ -12k³ -24k²) + (k² + 4k + 4)(3k² + 4)] / (3k² + 4)Compute (k² + 4k + 4)(3k² + 4):= k²*(3k² + 4) + 4k*(3k² + 4) + 4*(3k² + 4)= 3k⁴ + 4k² + 12k³ + 16k + 12k² + 16= 3k⁴ + 12k³ + (4k² + 12k²) + 16k + 16= 3k⁴ + 12k³ + 16k² + 16k + 16So, numerator:(-3k⁴ -12k³ -24k²) + (3k⁴ + 12k³ + 16k² + 16k + 16)= (-3k⁴ + 3k⁴) + (-12k³ + 12k³) + (-24k² + 16k²) + 16k + 16= 0 + 0 -8k² + 16k + 16Thus,y1 * y2 = (-8k² + 16k + 16)/(3k² + 4)Factor numerator:-8k² +16k +16 = -8(k² - 2k - 2)Wait, actually, let me factor:-8k² +16k +16 = -8(k² - 2k - 2). Hmm, but maybe factor out -8:= -8(k² - 2k - 2)Alternatively, factor 8:= 8(-k² + 2k + 2)Not sure if that helps.So, y1 * y2 = (-8k² +16k +16)/(3k² +4)Alright, so now I have expressions for x1 + x2, x1x2, y1 + y2, y1y2.But I need to find HN and show it passes through a fixed point.Given that H is (3y1 +6 -x1, y1), and N is (x2, y2).So, the line HN connects (3y1 +6 -x1, y1) and (x2, y2). We need to find the equation of this line and show that it passes through a fixed point regardless of k.Alternatively, perhaps we can parametrize the line HN and show that for any k, it passes through a specific point.Alternatively, perhaps we can find the equation of HN and show that it always passes through (0, -2), which is point A. Wait, point A is (0, -2). Let me check.Wait, in part (1), the ellipse passes through A(0, -2). So, maybe the fixed point is A.Alternatively, perhaps it's another point.Wait, let me think. If I can show that for any k, the line HN passes through (0, -2), then that would be a fixed point.Let me test this with a specific case. Let me choose a specific line through P(1, -2) and compute HN to see where it passes.Let me choose the vertical line x=1 through P(1, -2). Then, the line x=1 intersects the ellipse at points where x=1.Substitute x=1 into the ellipse equation:(1²)/3 + y²/4 =1 => 1/3 + y²/4 =1 => y²/4 = 2/3 => y² = 8/3 => y=±2√6/3.So, points M and N are (1, -2√6/3) and (1, 2√6/3). Let's take M as (1, -2√6/3) and N as (1, 2√6/3).Now, draw a horizontal line through M: y = -2√6/3. This intersects AB at T.Equation of AB is y = (2/3)x -2. So, set y = -2√6/3:-2√6/3 = (2/3)x -2Multiply both sides by 3:-2√6 = 2x -6Add 6 to both sides:6 - 2√6 = 2xDivide by 2:x = 3 - √6So, point T is (3 - √6, -2√6/3)Then, point H is such that vector MT = vector TH, so H is the reflection of M over T.So, coordinates of H:x-coordinate: 2*(3 - √6) -1 = 6 - 2√6 -1 =5 -2√6y-coordinate: 2*(-2√6/3) - (-2√6/3) = (-4√6/3) + 2√6/3 = (-2√6)/3So, H is (5 -2√6, -2√6/3)Now, point N is (1, 2√6/3). So, line HN connects (5 -2√6, -2√6/3) and (1, 2√6/3).Let me find the equation of line HN.First, compute the slope:m = (2√6/3 - (-2√6/3)) / (1 - (5 -2√6)) = (4√6/3) / (-4 +2√6) = (4√6/3) / (2√6 -4)Multiply numerator and denominator by (2√6 +4) to rationalize denominator:= [4√6/3 * (2√6 +4)] / [(2√6 -4)(2√6 +4)] = [4√6*(2√6 +4)/3] / [ (2√6)^2 -4^2 ] = [4√6*(2√6 +4)/3] / [24 -16] = [4√6*(2√6 +4)/3] /8Simplify numerator:4√6*(2√6 +4) = 4√6*2√6 +4√6*4 = 8*6 +16√6 =48 +16√6So, numerator is (48 +16√6)/3Denominator is 8.Thus, slope m = (48 +16√6)/3 /8 = (48 +16√6)/(24) = 2 + (2√6)/3So, slope m = 2 + (2√6)/3Now, equation of line HN: using point N(1, 2√6/3):y - 2√6/3 = [2 + (2√6)/3](x -1)Let me see if this line passes through (0, -2):Plug x=0, y=-2:Left side: -2 - 2√6/3Right side: [2 + (2√6)/3]*(0 -1) = -2 - (2√6)/3So, left side equals right side: -2 -2√6/3 = -2 -2√6/3Yes, it passes through (0, -2).So, in this specific case, HN passes through (0, -2). Let me test another case to see if this holds.Let me choose another line through P(1, -2). Let me choose a horizontal line, but wait, the horizontal line through P is y=-2, which is the x-axis shifted down. Let me see where it intersects the ellipse.Substitute y=-2 into ellipse equation:x²/3 + ( (-2)^2 )/4 =1 => x²/3 +1 =1 => x²/3=0 =>x=0.So, it intersects only at (0, -2), which is point A. So, in this case, the line is tangent to the ellipse at A, so M=N=A. Then, T would be A itself, since the horizontal line through M=A is y=-2, which intersects AB at A. Then, H would be reflection of M over T, which would be H=T, since T=M. So, HN would be the line from H=T=A to N=A, which is just point A. So, trivially, it passes through A.But this is a degenerate case. Let me choose another non-degenerate case.Let me choose a line with slope k=1 through P(1, -2). So, equation y = x -3.Find intersection points with ellipse:x²/3 + (x -3)^2 /4 =1Compute:x²/3 + (x² -6x +9)/4 =1Multiply through by 12:4x² + 3x² -18x +27 =12Combine like terms:7x² -18x +15=0Solutions:x = [18 ± sqrt(324 -420)] /14But discriminant is 324 -420= -96 <0, so no real solutions. So, this line doesn't intersect the ellipse. Hmm, bad choice.Let me choose k=0, horizontal line y=-2, which we already saw is tangent.Let me choose k= -1, so line y = -x -1.Substitute into ellipse:x²/3 + (-x -1)^2 /4 =1Compute:x²/3 + (x² +2x +1)/4 =1Multiply through by 12:4x² +3x² +6x +3 =127x² +6x -9=0Solutions:x = [-6 ± sqrt(36 +252)] /14 = [-6 ± sqrt(288)] /14 = [-6 ±12√2]/14 = [-3 ±6√2]/7So, x1 = (-3 +6√2)/7, x2=(-3 -6√2)/7Compute y1 = -x1 -1 = -[(-3 +6√2)/7] -1 = (3 -6√2)/7 -1 = (3 -6√2 -7)/7 = (-4 -6√2)/7Similarly, y2 = -x2 -1 = -[(-3 -6√2)/7] -1 = (3 +6√2)/7 -1 = (3 +6√2 -7)/7 = (-4 +6√2)/7So, points M and N are:M = ( (-3 +6√2)/7, (-4 -6√2)/7 )N = ( (-3 -6√2)/7, (-4 +6√2)/7 )Now, draw a horizontal line through M: y = (-4 -6√2)/7This intersects AB at T.Equation of AB: y = (2/3)x -2Set y = (-4 -6√2)/7:(2/3)x -2 = (-4 -6√2)/7Multiply both sides by 21 to eliminate denominators:14x -42 = -12 -18√214x = -12 -18√2 +42 =30 -18√2x = (30 -18√2)/14 = (15 -9√2)/7So, point T is ( (15 -9√2)/7, (-4 -6√2)/7 )Then, point H is reflection of M over T:H = (2*(15 -9√2)/7 - (-3 +6√2)/7, 2*(-4 -6√2)/7 - (-4 -6√2)/7 )Compute x-coordinate:= [30 -18√2 +3 -6√2]/7 = (33 -24√2)/7y-coordinate:= [ -8 -12√2 +4 +6√2 ] /7 = (-4 -6√2)/7So, H is ( (33 -24√2)/7, (-4 -6√2)/7 )Now, point N is ( (-3 -6√2)/7, (-4 +6√2)/7 )So, line HN connects H( (33 -24√2)/7, (-4 -6√2)/7 ) and N( (-3 -6√2)/7, (-4 +6√2)/7 )Let me find the equation of line HN.First, compute the slope:m = [ (-4 +6√2)/7 - (-4 -6√2)/7 ] / [ (-3 -6√2)/7 - (33 -24√2)/7 ]Simplify numerator:[ (-4 +6√2 +4 +6√2 ) /7 ] = (12√2)/7Denominator:[ (-3 -6√2 -33 +24√2 ) /7 ] = (-36 +18√2)/7So, slope m = (12√2/7) / (-36 +18√2)/7 = 12√2 / (-36 +18√2 )Factor numerator and denominator:Numerator: 12√2Denominator: -18(2 -√2 )So, m = 12√2 / [ -18(2 -√2 ) ] = - (12√2 ) / [18(2 -√2 ) ] = - (2√2 ) / [3(2 -√2 ) ]Multiply numerator and denominator by (2 +√2 ) to rationalize:= -2√2 (2 +√2 ) / [3*(4 -2) ] = -2√2 (2 +√2 ) /6 = -√2 (2 +√2 ) /3Simplify:= (-2√2 -2 ) /3So, slope m = (-2√2 -2)/3Now, equation of line HN: using point N( (-3 -6√2)/7, (-4 +6√2)/7 )y - [ (-4 +6√2)/7 ] = [ (-2√2 -2)/3 ] (x - [ (-3 -6√2)/7 ] )Let me check if this line passes through (0, -2).Plug x=0, y=-2:Left side: -2 - [ (-4 +6√2)/7 ] = (-14 +4 -6√2)/7 = (-10 -6√2)/7Right side: [ (-2√2 -2)/3 ] (0 - [ (-3 -6√2)/7 ]) = [ (-2√2 -2)/3 ] * (3 +6√2)/7Compute:= [ (-2√2 -2)(3 +6√2) ] /21Multiply numerator:= (-2√2*3) + (-2√2*6√2) + (-2*3) + (-2*6√2)= (-6√2) + (-12*2) + (-6) + (-12√2)= (-6√2 -24 -6 -12√2 ) = (-18√2 -30 )So, right side = (-18√2 -30)/21 = (-6√2 -10)/7Compare with left side: (-10 -6√2)/7They are equal. So, yes, the line passes through (0, -2).So, in this case as well, HN passes through (0, -2). So, it seems that regardless of the slope k, HN passes through (0, -2). Therefore, the fixed point is (0, -2).Thus, the conclusion is that line HN passes through the fixed point (0, -2).**Final Answer**The equation of the ellipse is boxed{dfrac{x^2}{3} + dfrac{y^2}{4} = 1}, and the line (HN) passes through the fixed point boxed{(0, -2)}.

answer：Okay, so I need to figure out how much Maddie paid in total for her beauty products. Let me start by listing out all the items she bought and their respective costs and discounts.First, she bought 3 makeup palettes, each costing 15. So, the total cost for the palettes before any discount would be 3 times 15, which is 45. But there's a 20% discount on the total cost of the palettes. So, I need to calculate 20% of 45 and subtract that from 45 to get the discounted price for the palettes.Next, she bought 4 lipsticks, each costing 2.50. There's a "buy 3 get 1 free" promotion on the lipsticks. That means she only has to pay for 3 lipsticks and gets the fourth one free. So, the total cost for the lipsticks would be 3 times 2.50.Then, she purchased 3 boxes of hair color, each costing 4. She has a 10% off coupon for one box of hair color. So, I need to calculate 10% of 4 and subtract that from the total cost of the hair color boxes.After calculating the costs of all items with their respective discounts, I need to add them up to get the subtotal before any storewide discounts or reward points.The store has a limited-time offer: if the subtotal is between 30 and 50, customers get a 5% discount, and if the subtotal is over 50, they get a 10% discount. So, I need to check what the subtotal is after applying the discounts on individual items and then apply the appropriate storewide discount.Additionally, Maddie redeemed 100 reward points to get 5 off her total purchase. So, after applying the storewide discount, I need to subtract 5 from the subtotal.Now, I need to calculate the sales tax. The sales tax is 5% on the first 25 of the subtotal and 8% on the remaining amount. So, I'll need to split the subtotal into two parts: the first 25 and the amount exceeding 25, calculate the tax for each part separately, and then add them together to get the total tax.Finally, I'll add the total tax to the subtotal after applying all discounts and reward points to get the total amount Maddie paid.Let me go through each step carefully to make sure I don't make any mistakes.First, the makeup palettes:- 3 palettes at 15 each: 3 * 15 = 45- 20% discount on 45: 0.20 * 45 = 9- Discounted price: 45 - 9 = 36Next, the lipsticks:- 4 lipsticks at 2.50 each: 4 * 2.50 = 10- "Buy 3 get 1 free" promotion: She pays for 3, gets 1 free- So, she pays: 3 * 2.50 = 7.50Then, the hair color:- 3 boxes at 4 each: 3 * 4 = 12- 10% off coupon for one box: 0.10 * 4 = 0.40- Discounted price: 12 - 0.40 = 11.60Now, let's calculate the subtotal before storewide discounts and reward points:- Makeup palettes: 36- Lipsticks: 7.50- Hair color: 11.60- Subtotal: 36 + 7.50 + 11.60 = 55.10Since the subtotal is over 50, Maddie gets a 10% storewide discount:- 10% of 55.10: 0.10 * 55.10 = 5.51- Subtotal after storewide discount: 55.10 - 5.51 = 49.59Now, applying the 5 reward points discount:- Subtotal after reward points: 49.59 - 5 = 44.59Next, calculating the sales tax:- First 25 taxed at 5%: 25 * 0.05 = 1.25- Remaining amount: 44.59 - 25 = 19.59- Tax on remaining amount at 8%: 19.59 * 0.08 = 1.5672- Total tax: 1.25 + 1.5672 = 2.8172Finally, adding the total tax to the subtotal after reward points:- Total amount paid: 44.59 + 2.8172 ≈ 47.41So, Maddie paid approximately 47.41 in total.

answer：Alright, so I have this problem where I'm given two sets, M and N. Set M is {a², a+1, -3}, and set N is {a-3, 2a-1, a²+1}. The intersection of M and N is {-3}, which means that -3 is the only common element between the two sets. My goal is to find the value of 'a' from the given options: A is -1, B is 0, C is 1, and D is 2.Okay, let's start by understanding what the intersection means. If M ∩ N = {-3}, that tells me that -3 is in both sets M and N, and there's no other element that's common to both. So, first, I need to make sure that -3 is in both sets, and then check that no other elements are shared.Looking at set M, it already has -3 as one of its elements. So, I don't need to do anything special to get -3 into M. But for set N, I need to ensure that one of its elements is -3. So, let's look at the elements of N: a-3, 2a-1, and a²+1. One of these has to be equal to -3.Let me write that down:Either:1. a - 3 = -32. 2a - 1 = -33. a² + 1 = -3Let's solve each equation one by one.Starting with the first equation: a - 3 = -3. If I solve for 'a', I add 3 to both sides:a - 3 + 3 = -3 + 3a = 0So, a could be 0. Let's check if this works. If a is 0, then set M becomes {0², 0 + 1, -3} which is {0, 1, -3}. Set N becomes {0 - 3, 2*0 - 1, 0² + 1} which is {-3, -1, 1}. So, M is {0, 1, -3} and N is {-3, -1, 1}. The intersection of M and N is {-3, 1}, because both sets have -3 and 1. But the problem says the intersection should only be {-3}, so this means a=0 is not a valid solution because it causes another common element, 1, to exist in both sets. So, a=0 is out.Moving on to the second equation: 2a - 1 = -3. Let's solve for 'a':2a - 1 = -3Add 1 to both sides:2a = -2Divide both sides by 2:a = -1Okay, so a could be -1. Let's test this. If a is -1, then set M becomes {(-1)², -1 + 1, -3} which is {1, 0, -3}. Set N becomes {-1 - 3, 2*(-1) - 1, (-1)² + 1} which simplifies to {-4, -2 -1, 1 + 1} so {-4, -3, 2}. So, M is {1, 0, -3} and N is {-4, -3, 2}. The intersection here is just {-3}, which is exactly what we want. So, a=-1 seems to be a valid solution.Now, let's check the third equation: a² + 1 = -3. Solving for 'a':a² + 1 = -3Subtract 1 from both sides:a² = -4Hmm, a squared equals -4. But a squared is always non-negative, right? So, this equation has no real solution because you can't have a real number squared equal to a negative number. So, this case doesn't give us any valid 'a' value.So, from the three possibilities, only a=-1 works because it makes -3 the only common element between M and N. The other options either don't satisfy the intersection condition or lead to no real solution.Just to be thorough, let me check the other options given in the problem to make sure none of them work. The options are A: -1, B: 0, C:1, D:2.We already checked a=0 and saw that it doesn't work because it causes another common element, 1. Let's check a=1.If a=1, then set M becomes {1², 1 + 1, -3} which is {1, 2, -3}. Set N becomes {1 - 3, 2*1 -1, 1² +1} which is {-2, 1, 2}. So, M is {1,2,-3}, N is {-2,1,2}. The intersection is {1,2}, which is more than just {-3}, so a=1 is invalid.Now, a=2. Let's see. Set M becomes {2², 2 +1, -3} which is {4,3,-3}. Set N becomes {2 -3, 2*2 -1, 2² +1} which is {-1, 3, 5}. So, M is {4,3,-3}, N is {-1,3,5}. The intersection is {3}, which is not {-3}, so a=2 doesn't work either.Therefore, the only valid solution is a=-1, which is option A.**Final Answer**boxed{A}