answer：Okay, so I have this problem about a parabola, a circle, and some lines and areas. Let me try to break it down step by step. First, part (1) is asking for the equation of the parabola G. It says the vertex is at the origin, and the focus is on the positive y-axis. So, I remember that for a parabola, the standard form when the vertex is at the origin and it opens upwards is ( y = frac{1}{4p}x^2 ), where p is the distance from the vertex to the focus. They also mention a point P(m, 4) whose distance to the directrix is 5. Since the focus is on the positive y-axis, the directrix should be a horizontal line below the vertex. If the focus is at (0, p), then the directrix is y = -p. The distance from point P(m, 4) to the directrix is given as 5. The distance from a point (x, y) to the line y = -p is just |y + p|. So, for point P(m, 4), the distance is |4 + p| = 5. This gives two possibilities: 4 + p = 5 or 4 + p = -5. Solving these, p = 1 or p = -9. But since the focus is on the positive y-axis, p must be positive, so p = 1. Therefore, the equation of the parabola is ( y = frac{1}{4}x^2 ). That seems straightforward. Moving on to part (2). It says a line passes through the focus of the parabola G and intersects the parabola and the circle ( x^2 + (y - 1)^2 = 1 ) at points A, C, D, B respectively. We need to prove that |AC||BD| is a constant value. Hmm, okay. The focus is at (0, 1), so any line passing through (0, 1) can be written in the form y = mx + 1, where m is the slope. Let me find the points of intersection between this line and the parabola. Substituting y = mx + 1 into the parabola equation ( y = frac{1}{4}x^2 ), we get:( mx + 1 = frac{1}{4}x^2 )Rearranging, we have:( frac{1}{4}x^2 - mx - 1 = 0 )Multiplying both sides by 4 to eliminate the fraction:( x^2 - 4mx - 4 = 0 )This is a quadratic equation in x. Let me denote the roots as x1 and x2. Using the quadratic formula:( x = frac{4m pm sqrt{(4m)^2 + 16}}{2} = frac{4m pm sqrt{16m^2 + 16}}{2} = frac{4m pm 4sqrt{m^2 + 1}}{2} = 2m pm 2sqrt{m^2 + 1} )So, the x-coordinates of points A and B are ( 2m + 2sqrt{m^2 + 1} ) and ( 2m - 2sqrt{m^2 + 1} ). Wait, but the line intersects the parabola at two points, so A and B are these points. But the problem mentions points A, C, D, B. So, maybe the line intersects the circle at points C and D as well. Let me find the intersection of the line y = mx + 1 with the circle ( x^2 + (y - 1)^2 = 1 ). Substituting y = mx + 1 into the circle equation:( x^2 + (mx + 1 - 1)^2 = 1 )( x^2 + (mx)^2 = 1 )( x^2(1 + m^2) = 1 )( x^2 = frac{1}{1 + m^2} )( x = pm frac{1}{sqrt{1 + m^2}} )So, the x-coordinates of points C and D are ( frac{1}{sqrt{1 + m^2}} ) and ( -frac{1}{sqrt{1 + m^2}} ). Therefore, the four points are:- A: ( (2m + 2sqrt{m^2 + 1}, m(2m + 2sqrt{m^2 + 1}) + 1) )- B: ( (2m - 2sqrt{m^2 + 1}, m(2m - 2sqrt{m^2 + 1}) + 1) )- C: ( left( frac{1}{sqrt{1 + m^2}}, m cdot frac{1}{sqrt{1 + m^2}} + 1 right) )- D: ( left( -frac{1}{sqrt{1 + m^2}}, -m cdot frac{1}{sqrt{1 + m^2}} + 1 right) )Wait, that seems complicated. Maybe there's a better way to compute |AC| and |BD| without getting into the coordinates directly. Alternatively, since the line passes through the focus, which is inside the circle, the line will intersect the circle at two points C and D, and the parabola at A and B. I recall that for a line intersecting a parabola and a circle, there might be some properties related to power of a point or something similar. But I'm not sure. Maybe using coordinates is the way to go. Let me try to compute |AC| and |BD|. First, let's find the coordinates of points A, B, C, D in terms of m.Points A and B on the parabola:- For point A: x = 2m + 2√(m² + 1), y = m*(2m + 2√(m² + 1)) + 1- For point B: x = 2m - 2√(m² + 1), y = m*(2m - 2√(m² + 1)) + 1Simplify y-coordinate for A:y = 2m² + 2m√(m² + 1) + 1Similarly, for B:y = 2m² - 2m√(m² + 1) + 1Points C and D on the circle:- Point C: x = 1/√(1 + m²), y = m/√(1 + m²) + 1- Point D: x = -1/√(1 + m²), y = -m/√(1 + m²) + 1Now, let's compute |AC| and |BD|.First, |AC| is the distance between A and C.Coordinates of A: (2m + 2√(m² + 1), 2m² + 2m√(m² + 1) + 1)Coordinates of C: (1/√(1 + m²), m/√(1 + m²) + 1)The distance |AC| is sqrt[(x_A - x_C)^2 + (y_A - y_C)^2]Similarly, |BD| is the distance between B and D.Coordinates of B: (2m - 2√(m² + 1), 2m² - 2m√(m² + 1) + 1)Coordinates of D: (-1/√(1 + m²), -m/√(1 + m²) + 1)Distance |BD| is sqrt[(x_B - x_D)^2 + (y_B - y_D)^2]This seems really messy. Maybe there's a pattern or simplification.Alternatively, maybe we can parametrize the line differently. Since the line passes through the focus (0,1), we can write it as y = t x + 1, where t is the slope. Wait, but I already did that with m. Maybe instead of m, use a parameter like t for the slope. Let me try that.Let me denote the slope as t, so the line is y = t x + 1.Intersection with the parabola y = (1/4)x²:t x + 1 = (1/4)x²Rearranged: (1/4)x² - t x - 1 = 0Multiply by 4: x² - 4 t x - 4 = 0Solutions: x = [4 t ± sqrt(16 t² + 16)] / 2 = 2 t ± 2 sqrt(t² + 1)So, points A and B have x-coordinates 2 t + 2 sqrt(t² + 1) and 2 t - 2 sqrt(t² + 1). Similarly, intersection with the circle x² + (y - 1)^2 = 1:Substitute y = t x + 1:x² + (t x + 1 - 1)^2 = 1x² + (t x)^2 = 1x² (1 + t²) = 1x² = 1 / (1 + t²)x = ±1 / sqrt(1 + t²)So, points C and D have x-coordinates 1 / sqrt(1 + t²) and -1 / sqrt(1 + t²). Now, let's compute |AC| and |BD|.First, |AC|:Point A: (2 t + 2 sqrt(t² + 1), y_A)Point C: (1 / sqrt(1 + t²), y_C)Compute y_A: y_A = t*(2 t + 2 sqrt(t² + 1)) + 1 = 2 t² + 2 t sqrt(t² + 1) + 1Compute y_C: y_C = t*(1 / sqrt(1 + t²)) + 1 = t / sqrt(1 + t²) + 1So, the difference in y-coordinates: y_A - y_C = 2 t² + 2 t sqrt(t² + 1) + 1 - (t / sqrt(1 + t²) + 1) = 2 t² + 2 t sqrt(t² + 1) - t / sqrt(1 + t²)Similarly, the difference in x-coordinates: x_A - x_C = 2 t + 2 sqrt(t² + 1) - 1 / sqrt(1 + t²)This is getting really complicated. Maybe instead of computing |AC| and |BD| directly, I can find |AC| * |BD|.Alternatively, maybe there's a property that |AC| * |BD| is constant regardless of the slope t.Wait, let me think about the power of a point. The power of point A with respect to the circle is |AC| * |AD|, but I'm not sure if that's directly applicable here.Alternatively, since the line passes through the focus, which is inside the circle, the product of the lengths from the focus to the points of intersection with the circle might be related.Wait, the circle is x² + (y - 1)^2 = 1, which is centered at (0,1) with radius 1. The focus is at (0,1), so it's the center of the circle. Therefore, any line through the center will intersect the circle at two points equidistant from the center.Wait, that's interesting. So, points C and D are symmetric with respect to the center (0,1). So, the distances from the center to C and D are both 1, but in opposite directions.But how does that relate to points A and B on the parabola?Alternatively, maybe we can use coordinates to compute |AC| * |BD|.But this seems too involved. Maybe there's a smarter way.Wait, let me consider the parametric equations. Let me set t as a parameter for the slope.Alternatively, maybe use parametric coordinates for the parabola.For the parabola y = (1/4)x², the parametric equations can be written as x = 2 t, y = t².So, any point on the parabola can be written as (2 t, t²). Then, the line passing through (0,1) and (2 t, t²) has slope m = (t² - 1)/(2 t - 0) = (t² - 1)/(2 t).So, the equation of the line is y = [(t² - 1)/(2 t)] x + 1.This line intersects the circle x² + (y - 1)^2 = 1.Substitute y = [(t² - 1)/(2 t)] x + 1 into the circle equation:x² + [((t² - 1)/(2 t)) x]^2 = 1Simplify:x² + [(t² - 1)^2 / (4 t²)] x² = 1Factor x²:x² [1 + (t² - 1)^2 / (4 t²)] = 1Compute the term inside the brackets:1 + (t^4 - 2 t² + 1) / (4 t²) = [4 t² + t^4 - 2 t² + 1] / (4 t²) = [t^4 + 2 t² + 1] / (4 t²) = (t² + 1)^2 / (4 t²)So, x² = 1 / [(t² + 1)^2 / (4 t²)] = 4 t² / (t² + 1)^2Thus, x = ± 2 t / (t² + 1)Therefore, the points C and D have x-coordinates 2 t / (t² + 1) and -2 t / (t² + 1). So, their coordinates are:C: (2 t / (t² + 1), y_C)D: (-2 t / (t² + 1), y_D)Compute y_C and y_D:y_C = [(t² - 1)/(2 t)] * (2 t / (t² + 1)) + 1 = (t² - 1)/(t² + 1) + 1 = [ (t² - 1) + (t² + 1) ] / (t² + 1) = (2 t²) / (t² + 1)Similarly, y_D = [(t² - 1)/(2 t)] * (-2 t / (t² + 1)) + 1 = - (t² - 1)/(t² + 1) + 1 = [ - (t² - 1) + (t² + 1) ] / (t² + 1) = (2) / (t² + 1)So, points C and D are:C: (2 t / (t² + 1), 2 t² / (t² + 1))D: (-2 t / (t² + 1), 2 / (t² + 1))Now, points A and B on the parabola are (2 t, t²) and another point. Wait, actually, when we parametrized the parabola as (2 t, t²), the line through (0,1) and (2 t, t²) intersects the parabola again at another point. Let me find that.The line intersects the parabola at two points: (2 t, t²) and another point. Let me find the other intersection.We have the line equation y = [(t² - 1)/(2 t)] x + 1.Set this equal to y = (1/4)x²:(1/4)x² = [(t² - 1)/(2 t)] x + 1Multiply both sides by 4:x² = 2 (t² - 1)/t x + 4Rearrange:x² - 2 (t² - 1)/t x - 4 = 0We know one root is x = 2 t, so let's factor it:(x - 2 t)(x + a) = 0Expanding: x² + (a - 2 t) x - 2 a t = 0Compare with original equation:x² - 2 (t² - 1)/t x - 4 = 0So, coefficients must match:a - 2 t = -2 (t² - 1)/t-2 a t = -4From the second equation: -2 a t = -4 => a = 2 / tSubstitute into first equation:2 / t - 2 t = -2 (t² - 1)/tMultiply both sides by t:2 - 2 t² = -2 (t² - 1)Simplify RHS: -2 t² + 2So, 2 - 2 t² = -2 t² + 2Which is an identity, so it checks out.Therefore, the other root is x = -a = -2 / tSo, the other intersection point B has x-coordinate -2 / t, and y-coordinate:y = (1/4)(-2 / t)^2 = (1/4)(4 / t²) = 1 / t²So, point B is (-2 / t, 1 / t²)Therefore, points A and B are:A: (2 t, t²)B: (-2 / t, 1 / t²)Now, points C and D are:C: (2 t / (t² + 1), 2 t² / (t² + 1))D: (-2 t / (t² + 1), 2 / (t² + 1))Now, let's compute |AC| and |BD|.First, |AC|:Coordinates of A: (2 t, t²)Coordinates of C: (2 t / (t² + 1), 2 t² / (t² + 1))Difference in x: 2 t - 2 t / (t² + 1) = 2 t [1 - 1 / (t² + 1)] = 2 t [ (t² + 1 - 1) / (t² + 1) ] = 2 t [ t² / (t² + 1) ] = 2 t³ / (t² + 1)Difference in y: t² - 2 t² / (t² + 1) = t² [1 - 2 / (t² + 1)] = t² [ (t² + 1 - 2) / (t² + 1) ] = t² [ (t² - 1) / (t² + 1) ] = t² (t² - 1) / (t² + 1)So, |AC| = sqrt[ (2 t³ / (t² + 1))² + (t² (t² - 1) / (t² + 1))² ]Factor out 1 / (t² + 1)^2:|AC| = sqrt[ (4 t^6 + t^4 (t² - 1)^2 ) / (t² + 1)^2 ]Compute numerator:4 t^6 + t^4 (t^4 - 2 t² + 1) = 4 t^6 + t^8 - 2 t^6 + t^4 = t^8 + 2 t^6 + t^4Factor numerator:t^4 (t^4 + 2 t² + 1) = t^4 (t² + 1)^2So, |AC| = sqrt[ t^4 (t² + 1)^2 / (t² + 1)^2 ] = sqrt(t^4) = t²Similarly, compute |BD|:Coordinates of B: (-2 / t, 1 / t²)Coordinates of D: (-2 t / (t² + 1), 2 / (t² + 1))Difference in x: (-2 / t) - (-2 t / (t² + 1)) = -2 / t + 2 t / (t² + 1) = (-2 (t² + 1) + 2 t²) / [t (t² + 1)] = (-2 t² - 2 + 2 t²) / [t (t² + 1)] = (-2) / [t (t² + 1)]Difference in y: (1 / t²) - (2 / (t² + 1)) = [ (t² + 1) - 2 t² ] / [ t² (t² + 1) ] = (1 - t²) / [ t² (t² + 1) ]So, |BD| = sqrt[ ( (-2) / [t (t² + 1)] )² + ( (1 - t²) / [ t² (t² + 1) ] )² ]Compute each term:First term: 4 / [ t² (t² + 1)^2 ]Second term: (1 - 2 t² + t^4) / [ t^4 (t² + 1)^2 ]Combine:[4 / t² + (1 - 2 t² + t^4) / t^4 ] / (t² + 1)^2Factor numerator:[4 t² + 1 - 2 t² + t^4] / t^4 = [ t^4 + 2 t² + 1 ] / t^4 = (t² + 1)^2 / t^4Thus, |BD| = sqrt[ (t² + 1)^2 / t^4 / (t² + 1)^2 ) ] = sqrt(1 / t^4 ) = 1 / t²Therefore, |AC| * |BD| = t² * (1 / t²) = 1So, |AC| * |BD| is a constant value of 1.Wow, that worked out nicely. I was worried it was going to be complicated, but by parametrizing the parabola and using the properties of the circle and the line through the focus, it simplified to a constant product.Now, moving on to part (3). We need to find the minimum sum of the areas of triangles ACM and BDM, where l1 and l2 are tangents to the parabola at points A and B, respectively, intersecting at point M.First, let's recall that the tangent to the parabola y = (1/4)x² at a point (x0, y0) is given by y = (x0 / 2) x - (x0² / 4). So, for point A: (2 t, t²), the tangent l1 is:y = (2 t / 2) x - ( (2 t)^2 / 4 ) = t x - t²Similarly, for point B: (-2 / t, 1 / t²), the tangent l2 is:y = ( (-2 / t) / 2 ) x - ( (-2 / t)^2 / 4 ) = (-1 / t) x - (4 / t²) / 4 = (-1 / t) x - (1 / t²)So, equations of l1 and l2 are:l1: y = t x - t²l2: y = (-1 / t) x - 1 / t²Find their intersection point M.Set t x - t² = (-1 / t) x - 1 / t²Bring all terms to one side:t x - t² + (1 / t) x + 1 / t² = 0Factor x:x (t + 1 / t) = t² - 1 / t²So,x = [ t² - 1 / t² ] / [ t + 1 / t ] = [ (t^4 - 1) / t² ] / [ (t² + 1) / t ] = [ (t^4 - 1) / t² ] * [ t / (t² + 1) ] = (t^4 - 1) / [ t (t² + 1) ]Factor numerator:t^4 - 1 = (t²)^2 - 1 = (t² - 1)(t² + 1)So,x = (t² - 1)(t² + 1) / [ t (t² + 1) ] = (t² - 1) / tTherefore, x = t - 1/tNow, substitute back into l1 to find y:y = t x - t² = t (t - 1/t) - t² = t² - 1 - t² = -1So, point M is ( t - 1/t, -1 )Now, we need to find the areas of triangles ACM and BDM.First, let's find the coordinates of all points:Points:A: (2 t, t²)C: (2 t / (t² + 1), 2 t² / (t² + 1))M: ( t - 1/t, -1 )Similarly,Points:B: (-2 / t, 1 / t²)D: (-2 t / (t² + 1), 2 / (t² + 1))M: ( t - 1/t, -1 )Compute area of triangle ACM.Using the formula for the area of a triangle given three points (x1,y1), (x2,y2), (x3,y3):Area = (1/2) | (x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1) |Alternatively, using determinant:Area = (1/2) | x_A(y_C - y_M) + x_C(y_M - y_A) + x_M(y_A - y_C) |Let me compute this.Compute for triangle ACM:x_A = 2 t, y_A = t²x_C = 2 t / (t² + 1), y_C = 2 t² / (t² + 1)x_M = t - 1/t, y_M = -1Compute:Area = (1/2) | x_A(y_C - y_M) + x_C(y_M - y_A) + x_M(y_A - y_C) |Plug in the values:= (1/2) | 2 t (2 t² / (t² + 1) - (-1)) + (2 t / (t² + 1)) (-1 - t²) + (t - 1/t)(t² - 2 t² / (t² + 1)) |Simplify term by term:First term: 2 t [ 2 t² / (t² + 1) + 1 ] = 2 t [ (2 t² + t² + 1) / (t² + 1) ] = 2 t [ (3 t² + 1) / (t² + 1) ] = 2 t (3 t² + 1) / (t² + 1)Second term: (2 t / (t² + 1)) [ -1 - t² ] = (2 t / (t² + 1)) [ - (t² + 1) ] = -2 tThird term: (t - 1/t) [ t² - 2 t² / (t² + 1) ] = (t - 1/t) [ (t² (t² + 1) - 2 t² ) / (t² + 1) ] = (t - 1/t) [ (t^4 + t² - 2 t² ) / (t² + 1) ] = (t - 1/t) [ (t^4 - t² ) / (t² + 1) ] = (t - 1/t) [ t² (t² - 1) / (t² + 1) ]Simplify:= [ t - 1/t ] * [ t² (t² - 1) / (t² + 1) ] = [ (t² - 1)/t ] * [ t² (t² - 1) / (t² + 1) ] = (t² - 1)^2 t / (t² + 1)So, putting it all together:Area_ACM = (1/2) | [ 2 t (3 t² + 1) / (t² + 1) ] + [ -2 t ] + [ (t² - 1)^2 t / (t² + 1) ] |Let me combine these terms:First term: 2 t (3 t² + 1) / (t² + 1)Second term: -2 tThird term: (t² - 1)^2 t / (t² + 1)Let me write all terms over the denominator (t² + 1):First term: 2 t (3 t² + 1)Second term: -2 t (t² + 1)Third term: (t² - 1)^2 tSo,Area_ACM = (1/2) | [ 2 t (3 t² + 1) - 2 t (t² + 1) + (t² - 1)^2 t ] / (t² + 1) |Simplify numerator:Expand each term:1. 2 t (3 t² + 1) = 6 t³ + 2 t2. -2 t (t² + 1) = -2 t³ - 2 t3. (t² - 1)^2 t = (t^4 - 2 t² + 1) t = t^5 - 2 t³ + tCombine all terms:6 t³ + 2 t - 2 t³ - 2 t + t^5 - 2 t³ + tSimplify:t^5 + (6 t³ - 2 t³ - 2 t³) + (2 t - 2 t + t) = t^5 + 2 t³ + tSo, numerator is t (t^4 + 2 t² + 1) = t (t² + 1)^2Thus,Area_ACM = (1/2) | [ t (t² + 1)^2 / (t² + 1) ] | = (1/2) | t (t² + 1) | = (1/2) t (t² + 1) since t is a real number (slope), but we can assume t > 0 for simplicity.Similarly, compute Area_BDM.Points:B: (-2 / t, 1 / t²)D: (-2 t / (t² + 1), 2 / (t² + 1))M: ( t - 1/t, -1 )Using the same area formula:Area = (1/2) | x_B(y_D - y_M) + x_D(y_M - y_B) + x_M(y_B - y_D) |Plug in the values:x_B = -2 / t, y_B = 1 / t²x_D = -2 t / (t² + 1), y_D = 2 / (t² + 1)x_M = t - 1/t, y_M = -1Compute:= (1/2) | (-2 / t)(2 / (t² + 1) - (-1)) + (-2 t / (t² + 1))(-1 - 1 / t²) + (t - 1/t)(1 / t² - 2 / (t² + 1)) |Simplify term by term:First term: (-2 / t) [ 2 / (t² + 1) + 1 ] = (-2 / t) [ (2 + t² + 1) / (t² + 1) ] = (-2 / t) [ (t² + 3) / (t² + 1) ] = -2 (t² + 3) / [ t (t² + 1) ]Second term: (-2 t / (t² + 1)) [ -1 - 1 / t² ] = (-2 t / (t² + 1)) [ - (t² + 1) / t² ] = (-2 t / (t² + 1)) * (- (t² + 1)/ t² ) = 2 t / t² = 2 / tThird term: (t - 1/t) [ 1 / t² - 2 / (t² + 1) ] = (t - 1/t) [ (t² + 1 - 2 t² ) / [ t² (t² + 1) ] ] = (t - 1/t) [ (1 - t² ) / [ t² (t² + 1) ] ] = [ (t² - 1)/t ] * [ (1 - t² ) / [ t² (t² + 1) ] ] = - (t² - 1)^2 / [ t^3 (t² + 1) ]So, putting it all together:Area_BDM = (1/2) | [ -2 (t² + 3) / [ t (t² + 1) ] + 2 / t - (t² - 1)^2 / [ t^3 (t² + 1) ] ] |Let me combine these terms over a common denominator, which is t^3 (t² + 1):First term: -2 (t² + 3) / [ t (t² + 1) ] = -2 (t² + 3) t² / [ t^3 (t² + 1) ] = -2 t² (t² + 3) / [ t^3 (t² + 1) ] = -2 (t² + 3) / [ t (t² + 1) ]Wait, maybe it's better to factor differently.Alternatively, let me compute each term:First term: -2 (t² + 3) / [ t (t² + 1) ]Second term: 2 / tThird term: - (t² - 1)^2 / [ t^3 (t² + 1) ]Let me write all terms over denominator t^3 (t² + 1):First term: -2 (t² + 3) * t² / [ t^3 (t² + 1) ] = -2 t² (t² + 3) / [ t^3 (t² + 1) ] = -2 (t² + 3) / [ t (t² + 1) ]Wait, that doesn't seem helpful. Maybe another approach.Alternatively, factor out 1 / [ t (t² + 1) ]:Area_BDM = (1/2) | [ -2 (t² + 3) + 2 (t² + 1) - (t² - 1)^2 / t² ] / [ t (t² + 1) ] |Wait, this is getting too convoluted. Maybe there's a symmetry or relation between Area_ACM and Area_BDM.Wait, from part (2), we found that |AC| * |BD| = 1. Maybe the areas have a similar relationship.Alternatively, notice that the expressions for Area_ACM and Area_BDM might be related through substitution of t with 1/t or something similar.Looking back at Area_ACM, we had:Area_ACM = (1/2) t (t² + 1)Similarly, let's see what Area_BDM would be.Wait, let me try to compute Area_BDM step by step.Compute the numerator:First term: -2 (t² + 3) / [ t (t² + 1) ]Second term: 2 / tThird term: - (t² - 1)^2 / [ t^3 (t² + 1) ]Let me combine the first and second terms:-2 (t² + 3) / [ t (t² + 1) ] + 2 / t = [ -2 (t² + 3) + 2 (t² + 1) ] / [ t (t² + 1) ] = [ -2 t² - 6 + 2 t² + 2 ] / [ t (t² + 1) ] = [ -4 ] / [ t (t² + 1) ]Now, add the third term:-4 / [ t (t² + 1) ] - (t² - 1)^2 / [ t^3 (t² + 1) ] = [ -4 t² - (t² - 1)^2 ] / [ t^3 (t² + 1) ]Expand (t² - 1)^2:= t^4 - 2 t² + 1So, numerator:-4 t² - t^4 + 2 t² - 1 = - t^4 - 2 t² - 1 = - (t^4 + 2 t² + 1) = - (t² + 1)^2Thus, Area_BDM = (1/2) | [ - (t² + 1)^2 / (t^3 (t² + 1)) ] | = (1/2) | - (t² + 1) / t^3 | = (1/2) (t² + 1) / t^3Therefore, Area_BDM = (1/2) (t² + 1) / t^3So, now, the total area is Area_ACM + Area_BDM = (1/2) t (t² + 1) + (1/2) (t² + 1)/t^3Factor out (1/2)(t² + 1):Total Area = (1/2)(t² + 1)( t + 1/t^3 )Simplify the expression inside:t + 1/t^3 = t + t^{-3}Let me write the total area as:Total Area = (1/2)(t² + 1)(t + 1/t^3 )Let me compute this expression:First, expand (t² + 1)(t + 1/t^3 ):= t² * t + t² * (1/t^3 ) + 1 * t + 1 * (1/t^3 )= t^3 + t^{-1} + t + t^{-3}So,Total Area = (1/2)( t^3 + t^{-1} + t + t^{-3} )Now, let me denote u = t + 1/t. Then, u² = t² + 2 + 1/t², and u³ = t³ + 3 t + 3 / t + 1 / t³.But our expression is t^3 + t^{-3} + t + t^{-1} = (t^3 + t^{-3}) + (t + t^{-1}) = (u³ - 3 u) + u = u³ - 2 uWait, let's check:u = t + 1/tu² = t² + 2 + 1/t²u³ = t³ + 3 t + 3 / t + 1/t³ = t³ + 1/t³ + 3 (t + 1/t ) = (t³ + 1/t³) + 3 uTherefore,t³ + 1/t³ = u³ - 3 uSo, our expression:t^3 + t^{-3} + t + t^{-1} = (u³ - 3 u) + u = u³ - 2 uTherefore,Total Area = (1/2)( u³ - 2 u ) = (1/2) u (u² - 2 )But u = t + 1/t, which is always greater than or equal to 2 for t > 0, by AM ≥ GM.So, u ≥ 2, with equality when t = 1.Now, we need to minimize Total Area = (1/2) u (u² - 2 )Let me denote f(u) = (1/2) u (u² - 2 ) = (1/2)(u³ - 2 u )Find the minimum of f(u) for u ≥ 2.Compute derivative f’(u):f’(u) = (1/2)(3 u² - 2 )Set derivative to zero:(1/2)(3 u² - 2 ) = 0 => 3 u² - 2 = 0 => u² = 2/3 => u = sqrt(2/3) ≈ 0.816But u ≥ 2, so the critical point is outside our domain.Therefore, the function f(u) is increasing for u ≥ 2, since f’(u) = (1/2)(3 u² - 2 ) > 0 when u² > 2/3, which is true for u ≥ 2.Thus, the minimum occurs at u = 2.Therefore, the minimum total area is f(2) = (1/2)(8 - 4 ) = (1/2)(4 ) = 2So, the minimum sum of the areas is 2.**Final Answer**(1) The equation of parabola ( G ) is (boxed{y = frac{1}{4}x^2}).(2) (|AC||BD|) is a constant value, specifically (boxed{1}).(3) The minimum sum of the areas of triangles (triangle ACM) and (triangle BDM) is (boxed{2}).

answer：First, I need to determine Mr. Bernard's current age based on the information provided. It is stated that in eight years, Mr. Bernard will be three times as old as Luke is now. Since Luke is currently 20 years old, I can calculate Mr. Bernard's age in eight years by multiplying Luke's current age by three.Next, I will find Mr. Bernard's current age by subtracting eight years from his age in eight years.With both Luke's and Mr. Bernard's current ages known, I can calculate their average age by adding their ages together and dividing by two.Finally, to find what is ten years less than their average age, I will subtract ten from the calculated average age.

answer：First, I need to determine how many ways I can choose two boxes out of the six to remain empty. This can be calculated using the combination formula ( binom{6}{2} ).Next, I have four distinct digits (1, 2, 3, and 4) that need to be placed in the remaining four boxes. Since each digit is unique and each box can hold only one digit, the number of ways to arrange these digits is given by the factorial of 4, which is ( 4! ).Finally, I multiply the number of ways to choose the empty boxes by the number of ways to arrange the digits to get the total number of distinct arrangements.

answer：First, I recognize that the given equation ( y^2 = 6x ) is in the standard form of a parabola that opens to the right, which is ( y^2 = 4px ).Next, I compare the two equations to identify the value of ( p ). By setting ( 4p = 6 ), I solve for ( p ) and find that ( p = frac{3}{2} ).Finally, knowing that the focus of such a parabola is located at ( (p, 0) ), I substitute ( p = frac{3}{2} ) to determine the coordinates of the focus.

answer：Okay, so I need to figure out how many hours Samantha sleeps every day. The problem says that today she woke up at 11 in the morning and went to bed at 7. Hmm, I'm not sure if the 7 is in the morning or evening. That's a bit confusing. Maybe I should assume it's 7 in the evening because people usually go to bed in the evening. Alright, if she went to bed at 7 PM and woke up at 11 AM the next day, I need to calculate the total hours she slept. Let me break it down. From 7 PM to midnight is 5 hours because 12 minus 7 is 5. Then, from midnight to 11 AM is 11 hours. So, adding those together, 5 plus 11 equals 16 hours. Wait, does that make sense? 16 hours seems like a lot for a person to sleep. Maybe I made a mistake. Let me double-check. If she went to bed at 7 PM and woke up at 11 AM, that's from 7 PM to 11 AM, which is indeed 16 hours. But maybe she didn't sleep the entire time. Perhaps she had some interruptions or she didn't sleep all the way through. Alternatively, if she went to bed at 7 AM and woke up at 11 AM, that would only be 4 hours, which seems too short. So, I think the first interpretation is correct—that she went to bed at 7 PM and slept until 11 AM, totaling 16 hours. I guess it's possible for someone to sleep that long, especially if they had a late night or needed extra rest. So, based on the information given, I think Samantha sleeps for 16 hours every day.

answer：First, I need to determine the discounted price of the milk. The original price is 4.00, and it's half off, so the discounted price is 2.00.Next, I'll calculate the cost of the bananas. At 0.75 per pound for 2 pounds, the total cost is 1.50.Then, I'll apply the coupon to the laundry detergent. The original price is 10.25, and with a 1.25 discount, the price becomes 9.00.Now, I'll add up the costs of all the items:- Milk: 2.00- Bread: 3.50- Laundry Detergent: 9.00- Bananas: 1.50The total cost is 2.00 + 3.50 + 9.00 + 1.50 = 16.00.Finally, I'll subtract the total cost from the 20 bill Josie was given to find out how much money she has left over:20.00 - 16.00 = 4.00.