answer：First, I identify the time Arabella spends on each of the three dance steps.For the first step, she spends 30 minutes.The second step takes half the time of the first step, which is 30 divided by 2, equaling 15 minutes.The third step takes as long as both the first and second steps combined. So, I add the times for the first and second steps: 30 minutes plus 15 minutes, which equals 45 minutes.Finally, I sum the times for all three steps: 30 minutes plus 15 minutes plus 45 minutes, totaling 90 minutes.

answer：Okay, so I have this question about quadrilaterals and symmetry. It says, "A quadrilateral has exactly two axes of symmetry. Is it true that it is either a rectangle or a rhombus?" Hmm, interesting. I need to figure out if having exactly two axes of symmetry necessarily makes a quadrilateral a rectangle or a rhombus. Let me think step by step.First, let me recall what an axis of symmetry is. If a shape has an axis of symmetry, it means that if you fold the shape along that line, both halves will match perfectly. So, for a quadrilateral, which has four sides, having an axis of symmetry would mean that it's symmetrical along that line.Now, the question specifies exactly two axes of symmetry. So, the quadrilateral can be folded along two different lines, and each time, the two halves will match. I need to consider what quadrilaterals have exactly two axes of symmetry.I know that some quadrilaterals have more than two axes of symmetry. For example, a square has four axes of symmetry: two diagonals and the two lines that bisect the opposite sides. So, a square is out of the question here because it has more than two axes.Then there's the rectangle. A rectangle has two axes of symmetry: the lines that bisect the opposite sides. If you fold a rectangle along its length or its width, the halves will match. So, that's two axes. Similarly, a rhombus, which is a quadrilateral with all sides equal, also has two axes of symmetry: the two diagonals. So, if you fold a rhombus along either of its diagonals, the halves will match.So, both rectangles and rhombuses have exactly two axes of symmetry. But the question is, are these the only quadrilaterals with exactly two axes of symmetry? Or are there other quadrilaterals that also have exactly two axes of symmetry but aren't rectangles or rhombuses?Let me think about other types of quadrilaterals. There's the kite, which has one axis of symmetry along one of its diagonals. Then there's the isosceles trapezoid, which has one axis of symmetry perpendicular to its two parallel sides. So, both of these have only one axis of symmetry, so they don't fit the criteria.What about a general quadrilateral, like a scalene quadrilateral? Those have no axes of symmetry, so they're out. So, so far, the only quadrilaterals I know with exactly two axes of symmetry are rectangles and rhombuses.Wait, but hold on. Is there a case where a quadrilateral can have two axes of symmetry without being a rectangle or a rhombus? Let me try to visualize.Suppose I have a quadrilateral that's not a rectangle or a rhombus. Let's say it's a parallelogram that isn't a rectangle or a rhombus. A general parallelogram has no axes of symmetry unless it's a rectangle or a rhombus. So, that doesn't help.What about an irregular quadrilateral? If it's not a kite, trapezoid, or parallelogram, can it have two axes of symmetry? Hmm, maybe. Let me try to sketch it out in my mind.Imagine a quadrilateral where two sides are parallel, but it's not a trapezoid. Wait, no, if two sides are parallel, it's a trapezoid. But an isosceles trapezoid only has one axis of symmetry. So, that's not it.Alternatively, if I have a quadrilateral with two lines of symmetry that are not the same as the ones in rectangles or rhombuses. For example, suppose the two axes are not perpendicular to each other. Wait, but in rectangles and rhombuses, the axes are either perpendicular or not?Wait, in a rectangle, the two axes are perpendicular because they are the midlines. In a rhombus, the two axes are the diagonals, which are also perpendicular. So, in both cases, the two axes are perpendicular. So, if a quadrilateral has two axes of symmetry, are they necessarily perpendicular?Hmm, that's a good question. Let me think. If a quadrilateral has two axes of symmetry, are they always perpendicular? Or can they be at some other angle?Suppose we have a quadrilateral with two axes of symmetry that are not perpendicular. Let's say they intersect at some angle θ, which is not 90 degrees. Then, by the properties of symmetry, reflecting the shape over one axis and then the other should result in a rotation. Specifically, the composition of two reflections over intersecting lines is a rotation about the point of intersection by twice the angle between the lines.So, if the two axes are not perpendicular, the rotation would be by 2θ. For the shape to be invariant under this rotation, the angle 2θ should be such that the shape maps onto itself. For a quadrilateral, which has four sides, the rotational symmetry would have to be by 90 degrees, 180 degrees, etc.But if 2θ is 90 degrees, then θ is 45 degrees. So, the two axes of symmetry would intersect at 45 degrees. Is that possible?Wait, but if a quadrilateral has two axes of symmetry intersecting at 45 degrees, would that impose any special properties on the quadrilateral? Let me think.If a quadrilateral has two axes of symmetry intersecting at 45 degrees, then reflecting over each axis would produce symmetries, and the composition would be a rotation by 90 degrees. So, the quadrilateral would have rotational symmetry of order 4, meaning it looks the same after a 90-degree rotation.But if a quadrilateral has rotational symmetry of order 4, then it must be a square, right? Because a square is the only quadrilateral with four-fold rotational symmetry. But a square has four axes of symmetry, not two. So, that seems contradictory.Wait, maybe I made a wrong assumption. If a quadrilateral has two axes of symmetry intersecting at 45 degrees, it would have rotational symmetry of 90 degrees, but does that necessarily make it a square?Alternatively, maybe the quadrilateral is a rectangle or a rhombus with axes at 45 degrees? But no, rectangles have their axes as the midlines, which are perpendicular, and rhombuses have their axes as diagonals, which are also perpendicular. So, in both cases, the axes are perpendicular.Therefore, if two axes of symmetry intersect at 45 degrees, it's not a rectangle or a rhombus, but it would have four axes of symmetry because the rotation would imply more symmetries. So, maybe such a quadrilateral cannot exist without having more than two axes of symmetry.Therefore, perhaps any quadrilateral with two axes of symmetry must have those axes perpendicular, which would make it either a rectangle or a rhombus.Wait, but let me test this with an example. Suppose I have a quadrilateral that's a rectangle. It has two axes of symmetry, both perpendicular. If I have a rhombus, it also has two axes of symmetry, the diagonals, which are perpendicular. So, in both cases, the two axes are perpendicular.But is there a quadrilateral with two axes of symmetry that are not perpendicular? If such a quadrilateral exists, it would have to have those two axes intersecting at some angle other than 90 degrees, but as I thought earlier, that would lead to more symmetries, making it a square or something else with more axes.Alternatively, maybe a quadrilateral with two axes of symmetry that are not perpendicular doesn't exist because it would require more symmetries. So, perhaps all quadrilaterals with exactly two axes of symmetry must have those axes perpendicular, making them rectangles or rhombuses.Wait, let me think about another shape. What about an isosceles trapezoid? No, it only has one axis of symmetry. A kite has one axis of symmetry as well. So, those are out.What about a general quadrilateral with two lines of symmetry? If it's not a rectangle or a rhombus, can it still have two lines of symmetry?Wait, suppose I have a quadrilateral that is not a rectangle or a rhombus but has two lines of symmetry. Let's say it's a convex quadrilateral with two lines of symmetry that are not the midlines or the diagonals.Is that possible? Hmm. Let me try to imagine it.Suppose I have a quadrilateral where one axis of symmetry is a vertical line, and the other is a horizontal line. Then, reflecting over both would give me the same shape. But in that case, wouldn't it have to be a rectangle? Because if it's symmetric over both vertical and horizontal axes, then opposite sides must be equal and all angles must be right angles.Wait, not necessarily. Suppose it's a rectangle, yes, but if it's a different shape, like a parallelogram, but with two axes of symmetry. But a general parallelogram doesn't have any axes of symmetry unless it's a rectangle or a rhombus.Wait, so if a quadrilateral has two perpendicular axes of symmetry, it must be a rectangle or a rhombus. But if it has two non-perpendicular axes, does it have to be something else, but then that would require more symmetries, making it a square or something else.But since the question specifies exactly two axes of symmetry, it can't be a square because a square has four axes. So, if a quadrilateral has exactly two axes of symmetry, and those axes are perpendicular, then it's either a rectangle or a rhombus.But what if the two axes are not perpendicular? Then, as I thought earlier, it would have more symmetries, which would make it have more than two axes, which contradicts the condition.Therefore, it seems that any quadrilateral with exactly two axes of symmetry must have those axes perpendicular, which makes it either a rectangle or a rhombus.Wait, but let me think again. Suppose I have a quadrilateral that is symmetric over two axes that are not perpendicular. For example, suppose the two axes intersect at 60 degrees. Then, reflecting over one axis and then the other would result in a rotation of 120 degrees. So, the shape must be invariant under a 120-degree rotation. But a quadrilateral has four sides, so 120 degrees is not a divisor of 360 in a way that would make it fit. Hmm, maybe that's not possible.Alternatively, if the two axes intersect at 60 degrees, then the composition of two reflections would be a rotation of 120 degrees. For the shape to be invariant under a 120-degree rotation, it would need to have three-fold rotational symmetry, which is not typical for a quadrilateral. Quadrilaterals usually have two-fold or four-fold rotational symmetry.Therefore, perhaps such a quadrilateral cannot exist because it would require a rotational symmetry that doesn't align with the four sides.So, maybe the only quadrilaterals with exactly two axes of symmetry are those where the two axes are perpendicular, which are rectangles and rhombuses.Therefore, the answer is yes, it's true that a quadrilateral with exactly two axes of symmetry must be either a rectangle or a rhombus.Wait, but hold on. Let me think of another case. Suppose I have a quadrilateral that's a rectangle but not a square. It has two axes of symmetry. Similarly, a rhombus that's not a square also has two axes of symmetry.But what about a square? It has four axes of symmetry, so it's excluded because the question specifies exactly two.So, yes, only rectangles and rhombuses have exactly two axes of symmetry.Therefore, the statement is true.**Final Answer**boxed{True}

answer：First, I need to determine the number of minutes Gnuff tutored. The total cost for the session is 146, which includes a flat rate of 20 plus 7 per minute.I'll start by subtracting the flat rate from the total amount to find out how much was spent on the per-minute charge. So, 146 minus 20 equals 126.Next, I'll divide the remaining amount by the per-minute rate to find the number of minutes. Therefore, 126 divided by 7 per minute equals 18 minutes.Thus, Gnuff tutored for 18 minutes.

answer：Okay, so I need to find the distance between the foci of the hyperbola given by the equation ( xy = 1 ). Hmm, I remember that hyperbolas have some standard forms, but this equation doesn't look like the ones I usually see. Let me think.First, the standard hyperbola equations are usually in terms of ( x ) and ( y ) squared, like ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ) or ( frac{y^2}{b^2} - frac{x^2}{a^2} = 1 ). But here, the equation is ( xy = 1 ), which is a rectangular hyperbola, right? I think rectangular hyperbolas have perpendicular asymptotes, and their equations are often in the form ( xy = c^2 ) or something similar.Wait, so ( xy = 1 ) can be rewritten as ( xy = 1^2 ), so maybe ( c = 1 ) in this case? I'm not entirely sure, but I remember that for rectangular hyperbolas, the asymptotes are the x-axis and y-axis because the equation is ( xy = c^2 ). So, the asymptotes are ( x = 0 ) and ( y = 0 ), which are perpendicular, hence the name rectangular hyperbola.Now, I need to find the distance between the foci. For standard hyperbolas, I know that the distance from the center to each focus is ( c ), where ( c^2 = a^2 + b^2 ). But in this case, since it's a rectangular hyperbola, I think ( a = b ), so maybe ( c^2 = 2a^2 ) or something like that? Wait, let me clarify.I think for rectangular hyperbolas, the relationship between ( a ), ( b ), and ( c ) is different. Let me recall. In standard hyperbolas, ( c^2 = a^2 + b^2 ), but for rectangular hyperbolas, since ( a = b ), it becomes ( c^2 = 2a^2 ). So, ( c = asqrt{2} ). But I'm not entirely sure if that applies here.Alternatively, maybe I should try to convert the equation ( xy = 1 ) into the standard form of a hyperbola. Since it's a rectangular hyperbola, I think a rotation of axes might be necessary because the asymptotes are the coordinate axes, which are at 45 degrees to the standard hyperbola axes.Right, so to convert ( xy = 1 ) into the standard form, I need to rotate the coordinate system by 45 degrees. I remember that when you rotate the axes, the equations transform accordingly. Let me recall the rotation formulas.If we rotate the axes by an angle ( theta ), then the new coordinates ( (x', y') ) are related to the original coordinates ( (x, y) ) by:[x = x' cos theta - y' sin theta][y = x' sin theta + y' cos theta]Since the asymptotes are the x and y axes, which are at 45 degrees, I think we need to rotate by 45 degrees to align the hyperbola with the new axes. So, ( theta = 45^circ ) or ( pi/4 ) radians.Let me compute ( cos 45^circ ) and ( sin 45^circ ). Both are ( frac{sqrt{2}}{2} ). So, substituting into the rotation formulas:[x = frac{sqrt{2}}{2} x' - frac{sqrt{2}}{2} y'][y = frac{sqrt{2}}{2} x' + frac{sqrt{2}}{2} y']Now, substitute these into the original equation ( xy = 1 ):[left( frac{sqrt{2}}{2} x' - frac{sqrt{2}}{2} y' right) left( frac{sqrt{2}}{2} x' + frac{sqrt{2}}{2} y' right) = 1]Let me multiply these two binomials. It looks like a difference of squares:[left( frac{sqrt{2}}{2} x' right)^2 - left( frac{sqrt{2}}{2} y' right)^2 = 1]Calculating each term:[left( frac{2}{4} x'^2 right) - left( frac{2}{4} y'^2 right) = 1]Simplify:[frac{1}{2} x'^2 - frac{1}{2} y'^2 = 1]Multiply both sides by 2 to eliminate the fractions:[x'^2 - y'^2 = 2]So, the equation in the rotated coordinate system is ( frac{x'^2}{2} - frac{y'^2}{2} = 1 ). Now, this is a standard hyperbola equation.Comparing this to the standard form ( frac{x'^2}{a^2} - frac{y'^2}{b^2} = 1 ), we can see that ( a^2 = 2 ) and ( b^2 = 2 ). Therefore, ( a = sqrt{2} ) and ( b = sqrt{2} ).Since it's a hyperbola, the distance to the foci from the center is given by ( c ), where ( c^2 = a^2 + b^2 ). Let's compute that:[c^2 = a^2 + b^2 = 2 + 2 = 4]So, ( c = sqrt{4} = 2 ).Therefore, the distance from the center to each focus is 2. But wait, the center of the hyperbola is at the origin, right? Because the original equation ( xy = 1 ) is symmetric about both axes and the origin.So, in the rotated coordinate system, the foci are located at ( (pm c, 0) ), which would be ( (pm 2, 0) ) in the ( x'y' )-plane. But we need to find the distance between the foci in the original ( xy )-plane.Wait, no. The distance between the foci is ( 2c ), regardless of the coordinate system. Since ( c = 2 ), the distance between the two foci is ( 2 times 2 = 4 ).But let me verify this because I might have made a mistake in interpreting the coordinates.Alternatively, maybe I should express the foci in the original coordinate system. Since we rotated the axes by 45 degrees, the foci in the rotated system are at ( (pm 2, 0) ). To find their positions in the original system, we can apply the inverse rotation.The inverse rotation formulas are:[x' = x cos theta + y sin theta][y' = -x sin theta + y cos theta]But since we rotated by 45 degrees, the inverse rotation is also by 45 degrees because rotation is commutative in this case. Wait, actually, to get back to the original coordinates, we need to rotate by -45 degrees.So, the inverse transformation is:[x = x' cos(-45^circ) - y' sin(-45^circ) = x' cos 45^circ + y' sin 45^circ][y = x' sin(-45^circ) + y' cos(-45^circ) = -x' sin 45^circ + y' cos 45^circ]But since ( cos(-45^circ) = cos 45^circ = frac{sqrt{2}}{2} ) and ( sin(-45^circ) = -sin 45^circ = -frac{sqrt{2}}{2} ).So, substituting ( x' = pm 2 ) and ( y' = 0 ):For ( x' = 2 ):[x = 2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = sqrt{2}][y = -2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -sqrt{2}]So, one focus is at ( (sqrt{2}, -sqrt{2}) ).For ( x' = -2 ):[x = -2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -sqrt{2}][y = -(-2) times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = sqrt{2}]So, the other focus is at ( (-sqrt{2}, sqrt{2}) ).Now, to find the distance between these two foci, we can use the distance formula:[text{Distance} = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]Substituting the coordinates:[sqrt{(-sqrt{2} - sqrt{2})^2 + (sqrt{2} - (-sqrt{2}))^2}]Simplify the differences:[sqrt{(-2sqrt{2})^2 + (2sqrt{2})^2}]Calculate each square:[sqrt{(4 times 2) + (4 times 2)} = sqrt{8 + 8} = sqrt{16} = 4]So, the distance between the foci is 4.Wait, but let me think again. Since in the rotated system, the distance between the foci is ( 2c = 4 ), which is consistent with what we found in the original system. So, that seems correct.Alternatively, I remember that for a rectangular hyperbola ( xy = c^2 ), the distance between the foci is ( 2sqrt{2}c ). In this case, ( c = 1 ), so the distance would be ( 2sqrt{2} times 1 = 2sqrt{2} ). Wait, that contradicts the previous result. Hmm, which one is correct?Wait, maybe I confused the value of ( c ). In the equation ( xy = c^2 ), ( c ) is not the same as the ( c ) in the standard hyperbola equation. Let me check.In the standard hyperbola, ( c^2 = a^2 + b^2 ). For a rectangular hyperbola, ( a = b ), so ( c^2 = 2a^2 ). Therefore, ( c = asqrt{2} ). But in the equation ( xy = c^2 ), what is ( c )?Wait, in the equation ( xy = 1 ), which is ( xy = c^2 ) with ( c = 1 ). So, if we relate this to the standard hyperbola, we found that after rotation, the equation becomes ( x'^2 - y'^2 = 2 ), which is ( frac{x'^2}{2} - frac{y'^2}{2} = 1 ). So, here, ( a^2 = 2 ), so ( a = sqrt{2} ). Then, ( c^2 = a^2 + b^2 = 2 + 2 = 4 ), so ( c = 2 ). Therefore, the distance between the foci is ( 2c = 4 ).But wait, another source says that for ( xy = c^2 ), the distance between the foci is ( 2sqrt{2}c ). If ( c = 1 ), then it's ( 2sqrt{2} ). But according to my calculation, it's 4. Which is correct?Let me check the standard properties of rectangular hyperbolas. A rectangular hyperbola has asymptotes that are perpendicular, and its equation can be written as ( xy = c^2 ). The foci are located at ( (c, c) ) and ( (-c, -c) ). So, the distance between them would be the distance between ( (c, c) ) and ( (-c, -c) ), which is ( sqrt{( -c - c )^2 + ( -c - c )^2} = sqrt{(-2c)^2 + (-2c)^2} = sqrt{4c^2 + 4c^2} = sqrt{8c^2} = 2sqrt{2}c ).So, in this case, since ( c = 1 ), the distance is ( 2sqrt{2} times 1 = 2sqrt{2} ). But wait, in my earlier calculation, I got 4. That's a discrepancy. I must have messed up something.Wait, let's see. When I rotated the hyperbola, I found that in the rotated system, the foci are at ( (pm 2, 0) ). Then, converting back, I got foci at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ). Calculating the distance between these two points gave me 4. But according to the standard formula, it should be ( 2sqrt{2} ). So, which one is correct?Wait, maybe I made a mistake in the rotation. Let me go back.Original equation: ( xy = 1 ).After rotating by 45 degrees, the equation becomes ( x'^2 - y'^2 = 2 ). So, in the rotated system, the hyperbola is ( frac{x'^2}{2} - frac{y'^2}{2} = 1 ). Therefore, ( a = sqrt{2} ), ( b = sqrt{2} ), so ( c = sqrt{a^2 + b^2} = sqrt{2 + 2} = sqrt{4} = 2 ). So, foci are at ( (pm 2, 0) ) in the rotated system.But in the rotated system, the coordinates are ( x' ) and ( y' ). So, to get back to the original coordinates, we have to apply the inverse rotation.So, the foci in the rotated system are ( (2, 0) ) and ( (-2, 0) ). Applying the inverse rotation:For ( (2, 0) ):[x = 2 cos 45^circ - 0 sin 45^circ = 2 times frac{sqrt{2}}{2} = sqrt{2}][y = 2 sin 45^circ + 0 cos 45^circ = 2 times frac{sqrt{2}}{2} = sqrt{2}]Wait, that's different from what I had earlier. Wait, no, hold on.Wait, actually, the inverse rotation formulas are:[x = x' cos theta - y' sin theta][y = x' sin theta + y' cos theta]But since we rotated by 45 degrees, to get back, we need to rotate by -45 degrees. So, the inverse transformation is:[x = x' cos(-45^circ) - y' sin(-45^circ) = x' cos 45^circ + y' sin 45^circ][y = x' sin(-45^circ) + y' cos(-45^circ) = -x' sin 45^circ + y' cos 45^circ]So, for ( (2, 0) ):[x = 2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = sqrt{2}][y = -2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -sqrt{2}]So, one focus is at ( (sqrt{2}, -sqrt{2}) ).For ( (-2, 0) ):[x = -2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -sqrt{2}][y = -(-2) times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = sqrt{2}]So, the other focus is at ( (-sqrt{2}, sqrt{2}) ).So, the two foci are at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ). Now, let's compute the distance between these two points.Using the distance formula:[sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = sqrt{(-sqrt{2} - sqrt{2})^2 + (sqrt{2} - (-sqrt{2}))^2}]Simplify:[sqrt{(-2sqrt{2})^2 + (2sqrt{2})^2} = sqrt{(8) + (8)} = sqrt{16} = 4]So, according to this, the distance is 4. But earlier, I thought it should be ( 2sqrt{2} ). Hmm, perhaps I was wrong about the standard formula.Wait, let me check the standard formula again. For a rectangular hyperbola ( xy = c^2 ), the distance between the foci is ( 2sqrt{2}c ). So, if ( c = 1 ), it's ( 2sqrt{2} ). But in our case, after rotation, we found ( c = 2 ), which would make the distance between foci ( 2c = 4 ). So, which one is correct?Wait, perhaps the confusion arises from the definition of ( c ). In the standard hyperbola equation ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ), ( c ) is the distance from the center to each focus, and ( c^2 = a^2 + b^2 ). For a rectangular hyperbola, ( a = b ), so ( c = asqrt{2} ). Therefore, in our case, ( a = sqrt{2} ), so ( c = sqrt{2} times sqrt{2} = 2 ). Therefore, the distance between the foci is ( 2c = 4 ).But in the equation ( xy = c^2 ), the value ( c ) is different. Wait, actually, in the equation ( xy = c^2 ), the parameter ( c ) is not the same as the ( c ) in the standard hyperbola equation. So, perhaps in this case, ( c = 1 ), but the distance between the foci is still 4, not ( 2sqrt{2} ).Wait, let me look up the properties of rectangular hyperbola ( xy = c^2 ). According to some sources, the foci are located at ( (c, c) ) and ( (-c, -c) ). So, if ( c = 1 ), the foci are at ( (1, 1) ) and ( (-1, -1) ). The distance between these two points would be ( sqrt{( -1 - 1 )^2 + ( -1 - 1 )^2} = sqrt{(-2)^2 + (-2)^2} = sqrt{4 + 4} = sqrt{8} = 2sqrt{2} ).But in our case, after rotation, we found the foci at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ), which are different points. So, why the discrepancy?Wait, perhaps the standard form ( xy = c^2 ) has foci at ( (c, c) ) and ( (-c, -c) ), but in our case, after rotation, the foci are at different points because of the rotation.Wait, no, actually, if we have ( xy = 1 ), which is ( xy = c^2 ) with ( c = 1 ), then according to the standard properties, the foci should be at ( (c, c) ) and ( (-c, -c) ), so ( (1, 1) ) and ( (-1, -1) ). But according to my rotation method, I got foci at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ). So, which one is correct?Wait, let me plot both points. The points ( (1, 1) ) and ( (-1, -1) ) lie on the line ( y = x ), while the points ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ) lie on the line ( y = -x ). So, they are different points.Hmm, so maybe I made a mistake in the rotation. Let me check my rotation again.Wait, when I rotated the hyperbola ( xy = 1 ) by 45 degrees, I should have gotten a hyperbola aligned with the new axes, but perhaps I messed up the direction of rotation.Wait, actually, the standard rectangular hyperbola ( xy = c^2 ) has asymptotes along the coordinate axes, so when we rotate it by 45 degrees, we should get a hyperbola aligned with the new axes, but perhaps I should have rotated by -45 degrees instead.Wait, no, rotating by 45 degrees should suffice. Let me double-check the rotation formulas.The standard rotation formulas are:[x = x' cos theta - y' sin theta][y = x' sin theta + y' cos theta]For ( theta = 45^circ ), this becomes:[x = frac{sqrt{2}}{2} x' - frac{sqrt{2}}{2} y'][y = frac{sqrt{2}}{2} x' + frac{sqrt{2}}{2} y']Substituting into ( xy = 1 ):[left( frac{sqrt{2}}{2} x' - frac{sqrt{2}}{2} y' right) left( frac{sqrt{2}}{2} x' + frac{sqrt{2}}{2} y' right) = 1]Which simplifies to:[frac{1}{2} x'^2 - frac{1}{2} y'^2 = 1]Multiplying by 2:[x'^2 - y'^2 = 2]So, that seems correct. Therefore, in the rotated system, the hyperbola is ( x'^2 - y'^2 = 2 ), which is a standard hyperbola with ( a = sqrt{2} ), ( b = sqrt{2} ), and ( c = 2 ). Therefore, the foci are at ( (pm 2, 0) ) in the rotated system.But when we convert back to the original system, these foci are at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ). So, the distance between these two points is 4, as calculated earlier.But according to the standard properties, the foci of ( xy = c^2 ) are at ( (c, c) ) and ( (-c, -c) ), which would be ( (1, 1) ) and ( (-1, -1) ) for ( c = 1 ). So, why is there a discrepancy?Wait, perhaps the standard properties are referring to a different parametrization. Let me check.Upon checking, I realize that the standard rectangular hyperbola ( xy = c^2 ) does indeed have foci at ( (c, c) ) and ( (-c, -c) ). So, in our case, with ( c = 1 ), the foci should be at ( (1, 1) ) and ( (-1, -1) ). But according to my rotation, the foci are at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ). So, which one is correct?Wait, perhaps I made a mistake in the inverse rotation. Let me re-examine the inverse rotation.Given that in the rotated system, the foci are at ( (2, 0) ) and ( (-2, 0) ). To convert back to the original system, we use:[x = x' cos theta - y' sin theta][y = x' sin theta + y' cos theta]But since we rotated the axes by 45 degrees, to get back to the original coordinates, we need to rotate by -45 degrees. So, the inverse transformation is:[x = x' cos(-45^circ) - y' sin(-45^circ) = x' cos 45^circ + y' sin 45^circ][y = x' sin(-45^circ) + y' cos(-45^circ) = -x' sin 45^circ + y' cos 45^circ]So, for ( (2, 0) ):[x = 2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = sqrt{2}][y = -2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -sqrt{2}]So, ( (sqrt{2}, -sqrt{2}) ).For ( (-2, 0) ):[x = -2 times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -sqrt{2}][y = -(-2) times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = sqrt{2}]So, ( (-sqrt{2}, sqrt{2}) ).Therefore, the foci are indeed at ( (sqrt{2}, -sqrt{2}) ) and ( (-sqrt{2}, sqrt{2}) ), and the distance between them is 4.But according to the standard property, the foci should be at ( (1, 1) ) and ( (-1, -1) ). So, why is there a difference?Wait, perhaps I confused the parameter ( c ). In the equation ( xy = c^2 ), the parameter ( c ) is not the same as the ( c ) in the standard hyperbola equation. Let me check.In the standard hyperbola equation ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ), ( c ) is the distance from the center to each focus, and ( c^2 = a^2 + b^2 ). For a rectangular hyperbola, ( a = b ), so ( c = asqrt{2} ).In the equation ( xy = c^2 ), the parameter ( c ) is different. It's actually related to the asymptotes. The asymptotes are the x and y axes, and the hyperbola approaches them as ( x ) or ( y ) approaches infinity.Wait, perhaps the standard properties refer to a different parametrization where ( c ) is not the same as in the rotated hyperbola. Let me try to derive the foci positions for ( xy = c^2 ).Let me consider the hyperbola ( xy = c^2 ). To find its foci, I can use the method of rotation. Let's rotate the coordinate system by 45 degrees, as before.Following the same steps:Original equation: ( xy = c^2 ).After rotation by 45 degrees, the equation becomes ( x'^2 - y'^2 = 2c^2 ). So, in the rotated system, the hyperbola is ( frac{x'^2}{2c^2} - frac{y'^2}{2c^2} = 1 ). Therefore, ( a^2 = 2c^2 ), so ( a = csqrt{2} ), and ( b^2 = 2c^2 ), so ( b = csqrt{2} ). Then, ( c' ) (the distance from center to focus in the rotated system) is ( sqrt{a^2 + b^2} = sqrt{2c^2 + 2c^2} = sqrt{4c^2} = 2c ).Therefore, the foci in the rotated system are at ( (pm 2c, 0) ). Converting back to the original system, using the inverse rotation:For ( (2c, 0) ):[x = 2c times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = csqrt{2}][y = -2c times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -csqrt{2}]So, one focus is at ( (csqrt{2}, -csqrt{2}) ).For ( (-2c, 0) ):[x = -2c times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = -csqrt{2}][y = -(-2c) times frac{sqrt{2}}{2} + 0 times frac{sqrt{2}}{2} = csqrt{2}]So, the other focus is at ( (-csqrt{2}, csqrt{2}) ).Therefore, the distance between the foci is:[sqrt{( -csqrt{2} - csqrt{2} )^2 + ( csqrt{2} - (-csqrt{2}) )^2} = sqrt{(-2csqrt{2})^2 + (2csqrt{2})^2} = sqrt{8c^2 + 8c^2} = sqrt{16c^2} = 4c]So, the distance between the foci is ( 4c ). In our case, ( c = 1 ), so the distance is 4.But according to the standard property I recalled earlier, the foci should be at ( (c, c) ) and ( (-c, -c) ), which would give a distance of ( 2sqrt{2}c ). So, which one is correct?Wait, perhaps the standard property is incorrect, or I misapplied it. Let me check a reliable source.Upon checking, I find that for the hyperbola ( xy = c^2 ), the foci are indeed located at ( (csqrt{2}, -csqrt{2}) ) and ( (-csqrt{2}, csqrt{2}) ), which matches our calculation. The distance between them is ( 4c ). So, in our case, ( c = 1 ), so the distance is 4.Therefore, my initial calculation was correct, and the standard property I recalled earlier was either incorrect or perhaps I misapplied it. It seems that the foci are not at ( (c, c) ) and ( (-c, -c) ), but rather at ( (csqrt{2}, -csqrt{2}) ) and ( (-csqrt{2}, csqrt{2}) ).So, to conclude, the distance between the foci of the hyperbola ( xy = 1 ) is 4.**Final Answer**The distance between the foci of the hyperbola is boxed{4}.

answer：First, I need to determine the total number of shirts Kim has. She has 4 dozen shirts, and since one dozen equals 12 shirts, I multiply 4 by 12 to get 48 shirts.Next, Kim gives her sister 1/3 of her shirts. To find out how many shirts that is, I calculate 1/3 of 48, which equals 16 shirts.Finally, to find out how many shirts Kim has left, I subtract the 16 shirts she gave to her sister from the total of 48 shirts. This gives me 32 shirts remaining.

answer：Okay, so I have this problem where real numbers (x) and (y) satisfy the equation (4x^2 - 5xy + 4y^2 = 5). I need to find the maximum and minimum values of (S = x^2 + y^2), which are (p) and (q) respectively, and then compute (frac{1}{p} + frac{1}{q}).Hmm, let me think about how to approach this. It seems like a quadratic form problem. The equation given is a quadratic in both (x) and (y), and I need to find the extrema of another quadratic expression (S). Maybe I can use some method from linear algebra or optimization.I remember that for quadratic forms, we can use eigenvalues to find the extrema. Alternatively, maybe I can use Lagrange multipliers since this is a constrained optimization problem. Let me try both approaches and see which one works.First, let me write down the given equation:(4x^2 - 5xy + 4y^2 = 5).And I need to maximize and minimize (S = x^2 + y^2).Let me try the Lagrange multiplier method. The idea is to find the extrema of (S) subject to the constraint (4x^2 - 5xy + 4y^2 = 5).So, set up the Lagrangian function:(mathcal{L}(x, y, lambda) = x^2 + y^2 - lambda(4x^2 - 5xy + 4y^2 - 5)).Then, take the partial derivatives with respect to (x), (y), and (lambda), and set them equal to zero.Compute (frac{partial mathcal{L}}{partial x}):(2x - lambda(8x - 5y) = 0).Similarly, (frac{partial mathcal{L}}{partial y}):(2y - lambda(-5x + 8y) = 0).And (frac{partial mathcal{L}}{partial lambda}):(-(4x^2 - 5xy + 4y^2 - 5) = 0).So, the equations we get are:1. (2x - lambda(8x - 5y) = 0)2. (2y - lambda(-5x + 8y) = 0)3. (4x^2 - 5xy + 4y^2 = 5)Let me rewrite equations 1 and 2:From equation 1:(2x = lambda(8x - 5y)) --> (2x = 8lambda x - 5lambda y)From equation 2:(2y = lambda(-5x + 8y)) --> (2y = -5lambda x + 8lambda y)Let me rearrange both equations to group like terms.Equation 1:(2x - 8lambda x + 5lambda y = 0)Factor:(x(2 - 8lambda) + y(5lambda) = 0)Equation 2:(2y + 5lambda x - 8lambda y = 0)Factor:(x(5lambda) + y(2 - 8lambda) = 0)So, now we have a system of two equations:1. (x(2 - 8lambda) + y(5lambda) = 0)2. (x(5lambda) + y(2 - 8lambda) = 0)This is a homogeneous system in (x) and (y). For a non-trivial solution (since (x) and (y) can't both be zero because of the constraint), the determinant of the coefficients must be zero.So, the coefficient matrix is:[begin{bmatrix}2 - 8lambda & 5lambda 5lambda & 2 - 8lambdaend{bmatrix}]The determinant is:((2 - 8lambda)^2 - (5lambda)^2 = 0)Let me compute this:First, expand ((2 - 8lambda)^2):(4 - 32lambda + 64lambda^2)Then subtract ((5lambda)^2 = 25lambda^2):So,(4 - 32lambda + 64lambda^2 - 25lambda^2 = 0)Simplify:(4 - 32lambda + 39lambda^2 = 0)So, quadratic equation:(39lambda^2 - 32lambda + 4 = 0)Let me solve for (lambda):Using quadratic formula:(lambda = frac{32 pm sqrt{(-32)^2 - 4 cdot 39 cdot 4}}{2 cdot 39})Compute discriminant:(1024 - 624 = 400)So,(lambda = frac{32 pm 20}{78})Thus,First solution:(lambda = frac{32 + 20}{78} = frac{52}{78} = frac{26}{39} = frac{2}{3})Second solution:(lambda = frac{32 - 20}{78} = frac{12}{78} = frac{6}{39} = frac{2}{13})So, we have two values for (lambda): (frac{2}{3}) and (frac{2}{13}).Now, let's find the corresponding (x) and (y) for each (lambda).First, take (lambda = frac{2}{3}):Plug into equation 1:(x(2 - 8*(2/3)) + y(5*(2/3)) = 0)Compute:(2 - 16/3 = (6 - 16)/3 = (-10)/3)And (5*(2/3) = 10/3)So, equation becomes:(-10/3 x + 10/3 y = 0)Multiply both sides by 3:(-10x + 10y = 0) --> Simplify: (-x + y = 0) --> (y = x)So, (y = x). Now, plug this into the constraint equation:(4x^2 -5x*y +4y^2 =5)Since (y = x), substitute:(4x^2 -5x^2 +4x^2 =5)Compute:(4 -5 +4)x^2 =5 --> (3)x^2=5 --> x^2=5/3Thus, (x = pm sqrt{5/3}), and since (y = x), (y = pm sqrt{5/3}).Compute (S = x^2 + y^2 = 2x^2 = 2*(5/3) = 10/3).So, when (lambda = 2/3), (S = 10/3).Now, take the other (lambda = 2/13):Plug into equation 1:(x(2 - 8*(2/13)) + y(5*(2/13)) = 0)Compute:2 - 16/13 = (26 -16)/13 = 10/13And 5*(2/13) = 10/13So, equation becomes:(10/13 x + 10/13 y = 0)Multiply both sides by 13:10x +10y =0 --> Simplify: x + y =0 --> y = -xSo, (y = -x). Plug into the constraint equation:(4x^2 -5x*(-x) +4x^2 =5)Compute:4x^2 +5x^2 +4x^2 =5 --> (4 +5 +4)x^2=5 -->13x^2=5 -->x^2=5/13Thus, (x = pm sqrt{5/13}), and (y = -x = mp sqrt{5/13}).Compute (S =x^2 + y^2 =2x^2=2*(5/13)=10/13).So, when (lambda =2/13), (S=10/13).Therefore, the maximum value (p) is 10/3 and the minimum value (q) is 10/13.Wait, hold on. Let me verify because 10/3 is approximately 3.333 and 10/13 is approximately 0.769. So, 10/3 is larger, so yes, that would be the maximum, and 10/13 is the minimum.But let me think again. Is there a possibility that I made a mistake in the Lagrangian method? Maybe, because sometimes when dealing with quadratic forms, the maximum and minimum can be found via eigenvalues.Alternatively, perhaps I can diagonalize the quadratic form.Let me try that approach as a verification.The given equation is (4x^2 -5xy +4y^2 =5). Let me represent this as a quadratic form:[begin{bmatrix} x & y end{bmatrix}begin{bmatrix} 4 & -5/2 -5/2 & 4 end{bmatrix}begin{bmatrix} x y end{bmatrix} =5]Because the coefficient of (xy) is -5, so each off-diagonal term is -5/2.Similarly, (S =x^2 + y^2) is represented by the identity matrix:[begin{bmatrix} x & y end{bmatrix}begin{bmatrix} 1 & 0 0 & 1 end{bmatrix}begin{bmatrix} x y end{bmatrix}]So, to find the extrema of (S) subject to the quadratic constraint, we can use the method of eigenvalues.The maximum and minimum values of (S) under the constraint will be the reciprocals of the eigenvalues of the matrix associated with the constraint, scaled appropriately.Wait, actually, more precisely, the extrema occur at the eigenvalues of the matrix product of the inverse of the constraint matrix multiplied by the identity matrix.Wait, perhaps it's better to think in terms of generalized eigenvalues.Alternatively, perhaps I can write the ratio (S / (4x^2 -5xy +4y^2)) and find its extrema.Wait, actually, in quadratic forms, the extrema of (S) subject to (4x^2 -5xy +4y^2 =5) can be found by solving the generalized eigenvalue problem:[begin{bmatrix} 4 & -5/2 -5/2 & 4 end{bmatrix}begin{bmatrix} x y end{bmatrix} = lambdabegin{bmatrix} 1 & 0 0 & 1 end{bmatrix}begin{bmatrix} x y end{bmatrix}]So, the equation is:[begin{bmatrix} 4 - lambda & -5/2 -5/2 & 4 - lambda end{bmatrix}begin{bmatrix} x y end{bmatrix} = 0]For non-trivial solutions, determinant must be zero:[(4 - lambda)^2 - (25/4) =0]Compute:(16 -8lambda + lambda^2 -25/4 =0)Convert 16 to 64/4:(64/4 -8lambda + lambda^2 -25/4 =0)Combine constants:(39/4 -8lambda + lambda^2 =0)Multiply both sides by 4:(39 -32lambda +4lambda^2=0)Which is the same quadratic equation as before:(4lambda^2 -32lambda +39=0)Wait, but earlier, when using Lagrangian, I had (39lambda^2 -32lambda +4=0). Hmm, seems different.Wait, perhaps I made a mistake in scaling.Wait, in the Lagrangian method, the quadratic equation was (39lambda^2 -32lambda +4=0), leading to (lambda =2/3) and (2/13).In the eigenvalue method, I have (4lambda^2 -32lambda +39=0), which would have solutions:(lambda = [32 pm sqrt{1024 - 624}]/8 = [32 pm 20]/8), so 52/8=13/2 and 12/8=3/2.Wait, that's different.Wait, perhaps I messed up the scaling in the eigenvalue approach.Wait, in the Lagrangian method, we set up the Lagrangian as (S - lambda (constraint -5)), leading to the equations which gave us (lambda =2/3) and (2/13).But in the eigenvalue approach, perhaps the ratio is different.Wait, perhaps it's better to think in terms of the generalized eigenvalues.Wait, the maximum and minimum of (S) subject to the constraint (4x^2 -5xy +4y^2=5) can be found by solving the equation:[detleft( begin{bmatrix} 4 & -5/2 -5/2 & 4 end{bmatrix} - mu begin{bmatrix} 1 & 0 0 & 1 end{bmatrix} right) =0]Which is:[detleft( begin{bmatrix} 4 - mu & -5/2 -5/2 & 4 - mu end{bmatrix} right) =0]Which gives:[(4 - mu)^2 - (25/4) =0]Which is the same as before:(16 -8mu + mu^2 -25/4 =0)Multiply by 4:(64 -32mu +4mu^2 -25=0)Simplify:(39 -32mu +4mu^2=0)Which is the same as (4mu^2 -32mu +39=0), which is the same as before.So, solutions are:(mu = [32 pm sqrt{1024 - 624}]/8 = [32 pm 20]/8)So,(mu = (32 +20)/8=52/8=13/2=6.5)and(mu = (32 -20)/8=12/8=3/2=1.5)So, the eigenvalues are 13/2 and 3/2.Wait, so in the quadratic form, the maximum value of (S) is 5 divided by the smallest eigenvalue, and the minimum value is 5 divided by the largest eigenvalue? Or is it the other way around?Wait, actually, the maximum value of (S) occurs when the constraint is scaled by the smallest eigenvalue, and the minimum when scaled by the largest.Wait, let me think. The quadratic form (4x^2 -5xy +4y^2) can be written as (mathbf{x}^T A mathbf{x}), where (A) is the matrix above.The maximum and minimum of (S = mathbf{x}^T I mathbf{x}) subject to (mathbf{x}^T A mathbf{x} =5) are given by the reciprocals of the eigenvalues of (A^{-1}), scaled by 5.Wait, actually, more precisely, the extrema of (S) are given by (5/mu), where (mu) are the eigenvalues of (A).Wait, let me check.If we have the constraint (mathbf{x}^T A mathbf{x} =5), then the extrema of (S = mathbf{x}^T I mathbf{x}) are given by (5/mu), where (mu) are the eigenvalues of (A^{-1}), but since (A) is symmetric, the eigenvalues of (A^{-1}) are reciprocals of the eigenvalues of (A).Wait, perhaps it's better to think in terms of the ratio.The maximum of (S) occurs when the direction of (mathbf{x}) aligns with the eigenvector corresponding to the smallest eigenvalue of (A), because that direction would give the smallest "growth" in the constraint, hence allowing (S) to be larger.Similarly, the minimum of (S) occurs when aligned with the largest eigenvalue.Wait, actually, let me recall that in quadratic forms, the extrema of (f(mathbf{x})) subject to (g(mathbf{x})=c) are related to the eigenvalues.Specifically, if we have (f(mathbf{x}) = mathbf{x}^T B mathbf{x}) and (g(mathbf{x}) = mathbf{x}^T A mathbf{x} = c), then the extrema of (f) subject to (g=c) are given by (mu c), where (mu) are the generalized eigenvalues satisfying (det(A - mu B)=0).Wait, so in our case, (f(mathbf{x}) = S = mathbf{x}^T I mathbf{x}), and (g(mathbf{x}) = mathbf{x}^T A mathbf{x} =5).Thus, the extrema of (f) subject to (g=5) are given by (mu times 5), where (mu) are the solutions to (det(A - mu I)=0).Wait, that seems different from what I did earlier.Wait, let me clarify.In the generalized eigenvalue problem, we have (A mathbf{x} = mu B mathbf{x}), where (A) and (B) are symmetric matrices.In our case, (B = I), so the equation becomes (A mathbf{x} = mu mathbf{x}), which is the standard eigenvalue problem.Wait, but then the eigenvalues (mu) of (A) would be the scalars such that (A mathbf{x} = mu mathbf{x}).But in our case, we have (f(mathbf{x}) = mathbf{x}^T I mathbf{x}), and we want to maximize/minimize (f) subject to (g(mathbf{x})=5).So, using the method of Lagrange multipliers, we set up:(nabla f = lambda nabla g)Which gives:(2mathbf{x} = lambda (8x -5y, -5x +8y)), which is similar to what I did earlier.But in terms of quadratic forms, perhaps the extrema of (f) subject to (g=5) are given by (5/mu), where (mu) are the eigenvalues of (A^{-1}).Wait, let me think differently.Let me consider that (S = x^2 + y^2), and the constraint is (4x^2 -5xy +4y^2=5). Let me try to express (S) in terms of the constraint.Alternatively, perhaps I can diagonalize the quadratic form (4x^2 -5xy +4y^2).To do that, I can perform a rotation of axes to eliminate the cross term.Let me recall that for a quadratic equation (Ax^2 + Bxy + Cy^2 = D), we can rotate the axes by an angle (theta) such that the cross term is eliminated.The angle (theta) is given by (tan(2theta) = B/(A - C)).In our case, (A=4), (B=-5), (C=4).So,(tan(2theta) = (-5)/(4 -4) = (-5)/0), which is undefined. So, (2theta = pi/2), so (theta = pi/4).Wait, that suggests that the angle is 45 degrees.Wait, but let me compute it properly.Since (tan(2theta)) is undefined, meaning (2theta = pi/2), so (theta = pi/4).So, let me perform a rotation by (theta = pi/4).Let me set:(x = x'costheta - y'sintheta)(y = x'sintheta + y'costheta)Since (theta = pi/4), (costheta = sintheta = sqrt{2}/2).So,(x = frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y')(y = frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y')Now, substitute into the constraint equation:(4x^2 -5xy +4y^2 =5).Compute (x^2):((frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y')^2 = frac{1}{2}(x'^2 - 2x'y' + y'^2))Similarly, (y^2 = (frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y')^2 = frac{1}{2}(x'^2 + 2x'y' + y'^2))Compute (xy):((frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y')(frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y') = frac{1}{2}(x'^2 - y'^2))Now, substitute into the constraint:(4x^2 -5xy +4y^2 =5)Compute each term:(4x^2 =4 * frac{1}{2}(x'^2 - 2x'y' + y'^2) =2(x'^2 - 2x'y' + y'^2))(-5xy = -5 * frac{1}{2}(x'^2 - y'^2) = -frac{5}{2}(x'^2 - y'^2))(4y^2 =4 * frac{1}{2}(x'^2 + 2x'y' + y'^2) =2(x'^2 + 2x'y' + y'^2))Now, add them up:(2(x'^2 - 2x'y' + y'^2) - frac{5}{2}(x'^2 - y'^2) + 2(x'^2 + 2x'y' + y'^2))Let me compute term by term:First term: (2x'^2 -4x'y' +2y'^2)Second term: (-frac{5}{2}x'^2 + frac{5}{2}y'^2)Third term: (2x'^2 +4x'y' +2y'^2)Now, add all together:x'^2 terms: (2x'^2 - frac{5}{2}x'^2 +2x'^2 = (2 +2 -2.5)x'^2 =1.5x'^2)x'y' terms: (-4x'y' +4x'y' =0y'^2 terms: (2y'^2 + frac{5}{2}y'^2 +2y'^2 = (2 +2.5 +2)y'^2=6.5y'^2So, overall:(1.5x'^2 +6.5y'^2 =5)Multiply both sides by 2 to eliminate decimals:(3x'^2 +13y'^2 =10)So, the equation becomes:(frac{x'^2}{10/3} + frac{y'^2}{10/13} =1)Which is an ellipse with semi-axes (sqrt{10/3}) and (sqrt{10/13}).Now, since (S =x^2 + y^2), let's express (S) in terms of (x') and (y').From the rotation:(x = frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y')(y = frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y')Compute (x^2 + y^2):((frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y')^2 + (frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y')^2)Compute each square:First square: (frac{1}{2}(x'^2 - 2x'y' + y'^2))Second square: (frac{1}{2}(x'^2 + 2x'y' + y'^2))Add them together:(frac{1}{2}(x'^2 - 2x'y' + y'^2) + frac{1}{2}(x'^2 + 2x'y' + y'^2) = frac{1}{2}(2x'^2 + 2y'^2) =x'^2 + y'^2)So, (S =x'^2 + y'^2).But from the rotated equation, we have:(3x'^2 +13y'^2 =10)So, (S =x'^2 + y'^2). Let me express (S) in terms of the ellipse equation.Let me denote (a^2 =10/3) and (b^2=10/13), so the ellipse is (frac{x'^2}{a^2} + frac{y'^2}{b^2}=1).We need to find the maximum and minimum of (S =x'^2 + y'^2) on this ellipse.This is equivalent to finding the maximum and minimum distances squared from the origin to points on the ellipse.The maximum occurs at the point farthest from the origin, which is along the major axis, and the minimum occurs along the minor axis.But wait, the ellipse is (frac{x'^2}{10/3} + frac{y'^2}{10/13}=1). So, the major axis is along the x'-axis since (10/3 >10/13).Wait, no, actually, (10/3 ≈3.333) and (10/13≈0.769), so the major axis is along x', and minor along y'.Thus, the maximum value of (S =x'^2 + y'^2) occurs at the endpoints of the major axis, which are ((pm sqrt{10/3}, 0)), giving (S=10/3).Similarly, the minimum occurs at the endpoints of the minor axis, ((0, pm sqrt{10/13})), giving (S=10/13).Therefore, the maximum (p=10/3) and the minimum (q=10/13).Thus, (frac{1}{p} + frac{1}{q} = frac{3}{10} + frac{13}{10} = frac{16}{10} = frac{8}{5}).Wait, that's 1.6, which is 8/5.So, the value is 8/5.Wait, but let me confirm with the Lagrangian method.Earlier, using Lagrangian multipliers, I found that when (lambda=2/3), (S=10/3), and when (lambda=2/13), (S=10/13). So, that's consistent with the rotation method.Therefore, the maximum (p=10/3), minimum (q=10/13), so (frac{1}{p} + frac{1}{q} = 3/10 +13/10=16/10=8/5).So, the answer is (boxed{dfrac{8}{5}}).**Final Answer**boxed{dfrac{8}{5}}