💡Okay, so I'm trying to find the standard equation of a hyperbola that passes through the point (4, √3) and has asymptotes y = ±(1/2)x. Hmm, hyperbolas can be a bit tricky, but let me see if I can work through this step by step.First, I remember that the standard form of a hyperbola depends on whether it opens horizontally or vertically. Since the asymptotes are y = ±(1/2)x, which are straight lines passing through the origin with slopes ±1/2, I think this hyperbola is centered at the origin. That makes things a bit simpler.Now, for a hyperbola centered at the origin, if it opens horizontally, its standard form is (x²/a²) - (y²/b²) = 1, and its asymptotes are y = ±(b/a)x. On the other hand, if it opens vertically, the standard form is (y²/a²) - (x²/b²) = 1, and the asymptotes are y = ±(a/b)x.Looking at the asymptotes given, y = ±(1/2)x, the slopes are ±1/2. Comparing this to the asymptote formulas, if it's a horizontal hyperbola, the slope would be ±(b/a), and if it's vertical, it would be ±(a/b). Let's test both possibilities.First, assuming it's a horizontal hyperbola: the slope is ±(b/a) = ±1/2. So, b/a = 1/2, which implies that b = a/2. That seems manageable.Alternatively, if it's a vertical hyperbola, the slope would be ±(a/b) = ±1/2, which would mean a/b = 1/2, so a = b/2. Hmm, that also seems possible.Wait, but I think the standard form for a horizontal hyperbola is more common when the slopes are less than 1, and vertical when slopes are greater than 1. Since 1/2 is less than 1, maybe it's a horizontal hyperbola. Let me go with that for now.So, assuming it's a horizontal hyperbola, the standard equation is (x²/a²) - (y²/b²) = 1, and we have b = a/2. So, I can write b in terms of a: b = a/2.Now, I need to find a and b such that the hyperbola passes through the point (4, √3). Let's plug that point into the equation.Plugging in x = 4 and y = √3:(4²/a²) - ((√3)²/b²) = 1Simplify:16/a² - 3/b² = 1But since b = a/2, let's substitute b in terms of a:16/a² - 3/( (a/2)² ) = 1Simplify the denominator in the second term:(a/2)² = a²/4, so 3/(a²/4) = 3 * (4/a²) = 12/a²So now, the equation becomes:16/a² - 12/a² = 1Combine like terms:(16 - 12)/a² = 14/a² = 1Multiply both sides by a²:4 = a²So, a² = 4, which means a = 2 (since a is positive in the standard form).Now, since b = a/2, and a = 2, then b = 2/2 = 1.So, b² = 1² = 1.Now, plug a² and b² back into the standard equation:(x²/4) - (y²/1) = 1Simplify:x²/4 - y² = 1Let me double-check if this hyperbola actually passes through the point (4, √3):Plug x = 4 and y = √3 into the equation:(4²)/4 - (√3)² = 16/4 - 3 = 4 - 3 = 1Yes, that works out. So, the equation x²/4 - y² = 1 is correct.Wait, just to be thorough, what if it was a vertical hyperbola? Let me check that possibility too.If it's a vertical hyperbola, the standard form is (y²/a²) - (x²/b²) = 1, and the asymptotes are y = ±(a/b)x. Given the asymptotes are y = ±(1/2)x, that would mean a/b = 1/2, so a = b/2.Let's try plugging in the point (4, √3) into this form:(y²/a²) - (x²/b²) = 1Substitute y = √3, x = 4:( (√3)² /a² ) - (4² / b² ) = 1Simplify:3/a² - 16/b² = 1But since a = b/2, then a² = (b²)/4. So, 3/(b²/4) - 16/b² = 1Simplify:3 * (4/b²) - 16/b² = 112/b² - 16/b² = 1Combine like terms:(12 - 16)/b² = 1-4/b² = 1Multiply both sides by b²:-4 = b²But b² can't be negative, so this doesn't make sense. Therefore, it must be a horizontal hyperbola.So, the standard equation is x²/4 - y² = 1.I think that's the answer. Let me just recap:1. Identified the hyperbola is centered at the origin because asymptotes pass through the origin.2. Determined the slopes of asymptotes to decide it's a horizontal hyperbola since slope is less than 1.3. Expressed b in terms of a using the slope.4. Plugged in the given point to find a² and b².5. Verified the equation with the given point.6. Checked the alternative case for thoroughness, which didn't work out.So, yeah, I'm confident the standard equation is x²/4 - y² = 1.

💡Okay, so I have this problem where point P(m, n) is moving on the parabola given by the equation y = -1/4 x². I need to find the minimum value of the expression sqrt(m² + (n + 1)²) + sqrt((m - 4)² + (n + 5)²). The options are A: 4, B: 5, C: sqrt(30), and D: 6.First, I should probably sketch the parabola to get a better understanding. The equation y = -1/4 x² is a downward-opening parabola because of the negative coefficient. The vertex is at the origin (0,0), and it's wider than the standard parabola y = -x² because of the 1/4 factor.Now, the expression I need to minimize is the sum of two square roots. Each square root represents the distance between point P(m, n) and two fixed points. Let me identify those points. The first square root, sqrt(m² + (n + 1)²), is the distance from P(m, n) to the point (0, -1). The second square root, sqrt((m - 4)² + (n + 5)²), is the distance from P(m, n) to the point (4, -5). So, essentially, I'm looking for the point P on the parabola that minimizes the sum of distances from P to (0, -1) and from P to (4, -5).This reminds me of the Fermat-Torricelli problem, which is about finding a point that minimizes the sum of distances to given points. However, in this case, the point P is constrained to lie on the parabola y = -1/4 x².Another thought: since the expression is the sum of distances from P to two fixed points, maybe I can use reflection properties of parabolas. I remember that one of the defining properties of a parabola is that any ray coming from the focus reflects off the parabola and travels parallel to the axis of symmetry. Conversely, any incoming ray parallel to the axis of symmetry reflects through the focus.Wait, so the focus of the parabola y = -1/4 x² is at (0, -1). Let me confirm that. The standard form of a parabola that opens downward is y = -1/(4p) x², where p is the distance from the vertex to the focus. Comparing this to y = -1/4 x², we have 1/(4p) = 1/4, so p = 1. Therefore, the focus is at (0, -1), which is one of the points in our distance expression. That's interesting.So, the first distance in the expression is from P to the focus F(0, -1). The second distance is from P to another point A(4, -5). So, we're trying to find the point P on the parabola such that the sum of the distances PF + PA is minimized.I recall that in optimization problems involving distances, especially with conic sections, reflection properties can be useful. For a parabola, the reflection property involves the focus and the directrix. The definition of a parabola is the set of all points equidistant from the focus and the directrix. The directrix of this parabola is the line y = 1 because for a parabola y = -1/(4p) x², the directrix is y = p, which in this case is y = 1.So, for any point P on the parabola, the distance from P to the focus F(0, -1) is equal to the distance from P to the directrix y = 1. That might be useful here.Let me denote the distance from P to F as d1 and the distance from P to A as d2. So, the expression we need to minimize is d1 + d2.Since d1 is equal to the distance from P to the directrix, which is a horizontal line y = 1, the distance from P(m, n) to the directrix is |n - 1|. But since the parabola is y = -1/4 x², n is always less than or equal to 0, so n - 1 is negative, making the distance 1 - n.Therefore, d1 = 1 - n.So, the expression becomes (1 - n) + sqrt((m - 4)² + (n + 5)²). Hmm, that might not immediately help, but perhaps I can express n in terms of m since P lies on the parabola.Given that P(m, n) is on the parabola y = -1/4 x², we have n = -1/4 m². So, I can substitute n in the expression.Let me write that out:Expression = sqrt(m² + (n + 1)²) + sqrt((m - 4)² + (n + 5)²)But n = -1/4 m², so substituting:= sqrt(m² + (-1/4 m² + 1)²) + sqrt((m - 4)² + (-1/4 m² + 5)²)This looks complicated, but maybe I can simplify it.Let me compute each term separately.First term: sqrt(m² + (-1/4 m² + 1)²)Let me compute (-1/4 m² + 1):= 1 - (1/4) m²So, the first term becomes sqrt(m² + (1 - (1/4)m²)²)Similarly, the second term is sqrt((m - 4)² + (-1/4 m² + 5)²)Compute (-1/4 m² + 5):= 5 - (1/4) m²So, the second term becomes sqrt((m - 4)² + (5 - (1/4)m²)²)So, now, the expression is:sqrt(m² + (1 - (1/4)m²)²) + sqrt((m - 4)² + (5 - (1/4)m²)²)This is still quite complex. Maybe I can consider calculus here, taking the derivative with respect to m and setting it to zero to find the minimum.But before diving into calculus, perhaps there's a geometric interpretation that can simplify this.Earlier, I noted that the first distance is from P to the focus F(0, -1), and the second distance is from P to A(4, -5). So, the expression is PF + PA.I remember that in optimization, sometimes reflecting a point across a line or curve can help find minimal paths. For example, in the reflection property of ellipses, the sum of distances from two foci is constant, but here it's a parabola, which only has one focus.Wait, but maybe I can use reflection across the directrix. Since the parabola is the set of points equidistant from the focus and the directrix, perhaps reflecting point A across the directrix could help.Let me try that. The directrix is y = 1. So, reflecting point A(4, -5) across the directrix y = 1.To reflect a point across a horizontal line y = k, the reflection of (x, y) is (x, 2k - y). So, reflecting A(4, -5) across y = 1:x-coordinate remains 4.y-coordinate becomes 2*1 - (-5) = 2 + 5 = 7.So, the reflection of A across the directrix is A'(4, 7).Now, what's the significance of this reflection? Well, for any point P on the parabola, the distance from P to F is equal to the distance from P to the directrix. So, PF = distance from P to directrix.But the distance from P to directrix is the same as the distance from P to A' minus something? Wait, maybe not directly.Alternatively, since PF = distance from P to directrix, which is 1 - n (as we had earlier), but I'm not sure that helps.Wait, another approach: since PF = distance from P to directrix, and the directrix is y = 1, perhaps the minimal path from F to A via a point P on the parabola is analogous to the reflection principle in mirrors, where the path taken is such that the angle of incidence equals the angle of reflection.But in this case, since the parabola reflects rays from the focus to be parallel, but here we have a different point A.Alternatively, perhaps the minimal sum PF + PA is equivalent to the distance from F to A', where A' is the reflection of A across the directrix.Wait, let me think. If I reflect A across the directrix to get A', then for any point P on the parabola, PF = distance from P to directrix, which is the same as the distance from P to A' minus something? Hmm, maybe not.Wait, another thought: since PF = distance from P to directrix, which is 1 - n, and PA is the distance from P to A(4, -5). So, PF + PA = (1 - n) + PA.But I don't see an immediate way to minimize this.Alternatively, perhaps I can consider that the minimal PF + PA is equivalent to the minimal distance from F to A via a point P on the parabola. So, if I can find a point P on the parabola such that the path from F to P to A is minimized, that would give me the minimal sum.But how?Wait, in the reflection principle, the shortest path from F to A via a point P on the parabola would involve reflecting A across the parabola's axis or something. But I'm not sure.Alternatively, since the parabola is the set of points equidistant from F and the directrix, maybe I can use that to rewrite the expression.Wait, PF = distance from P to directrix, which is 1 - n. So, PF + PA = (1 - n) + PA.So, the expression becomes (1 - n) + PA.But n = -1/4 m², so 1 - n = 1 + 1/4 m².So, the expression is 1 + 1/4 m² + sqrt((m - 4)² + (n + 5)²).But n = -1/4 m², so n + 5 = 5 - 1/4 m².So, the second term is sqrt((m - 4)² + (5 - 1/4 m²)²).So, the expression is:1 + (1/4)m² + sqrt((m - 4)² + (5 - (1/4)m²)²)This still looks complicated, but maybe I can consider taking the derivative with respect to m and setting it to zero.Let me denote f(m) = 1 + (1/4)m² + sqrt((m - 4)² + (5 - (1/4)m²)²)To find the minimum, compute f'(m) and set it to zero.First, compute the derivative of the first two terms:d/dm [1 + (1/4)m²] = (1/2)mNow, for the square root term, let me denote:g(m) = sqrt((m - 4)² + (5 - (1/4)m²)²)So, g(m) = sqrt[(m - 4)² + (5 - (1/4)m²)²]Compute g'(m):g'(m) = [1/(2g(m))] * [2(m - 4) + 2(5 - (1/4)m²)(- (1/2)m)]Simplify:= [ (m - 4) + (5 - (1/4)m²)(- (1/2)m) ] / g(m)So, putting it all together, f'(m) = (1/2)m + [ (m - 4) - (1/2)m(5 - (1/4)m²) ] / g(m)Set f'(m) = 0:(1/2)m + [ (m - 4) - (1/2)m(5 - (1/4)m²) ] / g(m) = 0This equation seems quite complicated. Maybe there's a better approach.Wait, going back to the reflection idea. If I reflect point A across the directrix, which is y = 1, to get A'(4, 7), as I did earlier. Then, the distance from P to A is equal to the distance from P to A' minus twice the distance from P to the directrix? Hmm, not sure.Wait, actually, reflecting A across the directrix gives A', and since PF = distance from P to directrix, which is 1 - n, perhaps the minimal PF + PA is equivalent to the minimal distance from F to A' via P on the parabola.Wait, maybe not. Let me think differently.If I consider that PF = distance from P to directrix, which is 1 - n, and PA is the distance from P to A. So, PF + PA = (1 - n) + PA.But since n = -1/4 m², we can write 1 - n = 1 + 1/4 m².So, PF + PA = 1 + 1/4 m² + PA.But PA is sqrt((m - 4)^2 + (n + 5)^2) = sqrt((m - 4)^2 + (5 - 1/4 m²)^2).So, PF + PA = 1 + 1/4 m² + sqrt((m - 4)^2 + (5 - 1/4 m²)^2).This still seems complicated, but maybe I can consider that the minimal value occurs when the derivative is zero, as I started earlier.Alternatively, perhaps I can consider that the minimal sum PF + PA is equal to the distance from F to A', where A' is the reflection of A across the directrix.Wait, let me test this idea.If I reflect A(4, -5) across the directrix y = 1, I get A'(4, 7). Then, the distance from F(0, -1) to A'(4, 7) is sqrt((4 - 0)^2 + (7 - (-1))^2) = sqrt(16 + 64) = sqrt(80) = 4 sqrt(5) ≈ 8.944.But the options given are 4, 5, sqrt(30) ≈ 5.477, and 6. So, 4 sqrt(5) is not among the options, so maybe this approach is not correct.Wait, perhaps I'm overcomplicating it. Let me consider the reflection property of the parabola. Since the parabola reflects rays from the focus to be parallel to the axis, but here we have another point A. Maybe the minimal path from F to A via P on the parabola is when the angle of incidence equals the angle of reflection with respect to the tangent at P.But I'm not sure how to apply that here.Alternatively, perhaps I can parametrize the parabola and then express the sum of distances in terms of a single variable, then take the derivative.Given that P(m, n) is on the parabola y = -1/4 x², so n = -1/4 m².So, let me write the expression as:sqrt(m² + (n + 1)^2) + sqrt((m - 4)^2 + (n + 5)^2)Substitute n = -1/4 m²:= sqrt(m² + (-1/4 m² + 1)^2) + sqrt((m - 4)^2 + (-1/4 m² + 5)^2)Let me compute each term:First term: sqrt(m² + (1 - (1/4)m²)^2)Let me expand (1 - (1/4)m²)^2:= 1 - (1/2)m² + (1/16)m⁴So, first term becomes sqrt(m² + 1 - (1/2)m² + (1/16)m⁴) = sqrt(1 + (1/2)m² + (1/16)m⁴)Similarly, second term: sqrt((m - 4)^2 + (5 - (1/4)m²)^2)Compute (5 - (1/4)m²)^2:= 25 - (5/2)m² + (1/16)m⁴Compute (m - 4)^2:= m² - 8m + 16So, second term becomes sqrt(m² - 8m + 16 + 25 - (5/2)m² + (1/16)m⁴) = sqrt(- (3/2)m² - 8m + 41 + (1/16)m⁴)So, the expression is:sqrt(1 + (1/2)m² + (1/16)m⁴) + sqrt(- (3/2)m² - 8m + 41 + (1/16)m⁴)This is quite messy. Maybe I can consider that the minimal value occurs when the derivative is zero, but this seems too involved.Alternatively, perhaps I can consider specific points on the parabola and see which one gives the minimal sum.Let me try m = 0:At m = 0, n = 0.First distance: sqrt(0 + (0 + 1)^2) = 1Second distance: sqrt((0 - 4)^2 + (0 + 5)^2) = sqrt(16 + 25) = sqrt(41) ≈ 6.403Total: 1 + 6.403 ≈ 7.403Not one of the options, but higher than 6.Next, try m = 2:n = -1/4*(4) = -1First distance: sqrt(4 + (-1 + 1)^2) = sqrt(4 + 0) = 2Second distance: sqrt((2 - 4)^2 + (-1 + 5)^2) = sqrt(4 + 16) = sqrt(20) ≈ 4.472Total: 2 + 4.472 ≈ 6.472Closer to 6, but still higher.Next, try m = 4:n = -1/4*(16) = -4First distance: sqrt(16 + (-4 + 1)^2) = sqrt(16 + 9) = 5Second distance: sqrt((4 - 4)^2 + (-4 + 5)^2) = sqrt(0 + 1) = 1Total: 5 + 1 = 6Hey, that's one of the options, D: 6.So, at m = 4, n = -4, the sum is 6.Is this the minimal value?Wait, let me check m = 3:n = -1/4*(9) = -2.25First distance: sqrt(9 + (-2.25 + 1)^2) = sqrt(9 + (-1.25)^2) = sqrt(9 + 1.5625) = sqrt(10.5625) ≈ 3.25Second distance: sqrt((3 - 4)^2 + (-2.25 + 5)^2) = sqrt(1 + (2.75)^2) = sqrt(1 + 7.5625) = sqrt(8.5625) ≈ 2.926Total: ≈ 3.25 + 2.926 ≈ 6.176So, higher than 6.What about m = 5:n = -1/4*(25) = -6.25First distance: sqrt(25 + (-6.25 + 1)^2) = sqrt(25 + (-5.25)^2) = sqrt(25 + 27.5625) = sqrt(52.5625) ≈ 7.25Second distance: sqrt((5 - 4)^2 + (-6.25 + 5)^2) = sqrt(1 + (-1.25)^2) = sqrt(1 + 1.5625) = sqrt(2.5625) = 1.600Total: ≈ 7.25 + 1.600 ≈ 8.85Higher.What about m = 1:n = -1/4*(1) = -0.25First distance: sqrt(1 + (-0.25 + 1)^2) = sqrt(1 + (0.75)^2) = sqrt(1 + 0.5625) = sqrt(1.5625) = 1.25Second distance: sqrt((1 - 4)^2 + (-0.25 + 5)^2) = sqrt(9 + (4.75)^2) = sqrt(9 + 22.5625) = sqrt(31.5625) ≈ 5.618Total: ≈ 1.25 + 5.618 ≈ 6.868Still higher than 6.What about m = -4:n = -1/4*(16) = -4First distance: sqrt(16 + (-4 + 1)^2) = sqrt(16 + 9) = 5Second distance: sqrt((-4 - 4)^2 + (-4 + 5)^2) = sqrt(64 + 1) = sqrt(65) ≈ 8.062Total: 5 + 8.062 ≈ 13.062Way higher.So, from these trials, the minimal value seems to occur at m = 4, giving a total of 6.But wait, let me check m = something else, say m = sqrt(16) = 4, which we already did.Alternatively, maybe m = something else gives a lower value.Wait, let me think about the reflection idea again. If I reflect point A across the directrix to get A'(4,7), then the minimal distance from F to A via P on the parabola would be the straight line distance from F to A', but since P must lie on the parabola, maybe the minimal sum is the distance from F to A'.Wait, distance from F(0, -1) to A'(4,7) is sqrt((4)^2 + (8)^2) = sqrt(16 + 64) = sqrt(80) ≈ 8.944, which is higher than 6, so that doesn't help.Wait, but earlier, when I took m =4, n=-4, the sum was 6, which is one of the options. So, maybe that's the minimal value.But let me confirm if this is indeed the minimal.Wait, when m =4, n = -4, which is on the parabola y = -1/4 x², since -1/4*(16) = -4.So, point P is (4, -4).Distance from P to F(0, -1): sqrt((4)^2 + (-4 +1)^2) = sqrt(16 + 9) = 5Distance from P to A(4, -5): sqrt((4-4)^2 + (-4 +5)^2) = sqrt(0 +1) =1Total: 5 +1=6So, that's correct.Is there a point P on the parabola where the sum is less than 6?From the trials above, it seems not. The next closest was m=2, giving about 6.472, which is higher.Alternatively, perhaps the minimal value is indeed 6, achieved at m=4, n=-4.Therefore, the answer is D:6.

💡Okay, so I have this problem about a circle in the Cartesian coordinate plane. Let me try to understand what it's asking.Part (I) says: A circle P intercepts a line segment of length 2√2 on the x-axis and a line segment of length 2√3 on the y-axis. I need to find the trajectory equation of the center of the circle P.Hmm, okay. So, the circle intersects the x-axis and y-axis, creating chords of those lengths. The center of the circle is moving such that these chord lengths remain constant. I need to find the equation that describes the path of the center.Let me recall some geometry. When a circle intersects an axis, the length of the chord can be related to the radius and the distance from the center to the axis.Suppose the center of the circle is at point (a, b). The radius of the circle is R. For the x-axis: The circle intersects the x-axis at two points, forming a chord of length 2√2. The distance from the center to the x-axis is |b|. Using the formula for the length of a chord, which is 2√(R² - d²), where d is the distance from the center to the axis.So, for the x-axis: 2√2 = 2√(R² - b²). Dividing both sides by 2: √2 = √(R² - b²). Squaring both sides: 2 = R² - b². So, R² - b² = 2.Similarly, for the y-axis: The chord length is 2√3. The distance from the center to the y-axis is |a|. So, 2√3 = 2√(R² - a²). Dividing by 2: √3 = √(R² - a²). Squaring: 3 = R² - a². So, R² - a² = 3.Now, I have two equations:1. R² - b² = 22. R² - a² = 3If I subtract the first equation from the second, I get:(R² - a²) - (R² - b²) = 3 - 2Simplify: -a² + b² = 1Which can be written as: b² - a² = 1So, the equation relating a and b is b² - a² = 1. Therefore, the trajectory of the center (a, b) is given by y² - x² = 1.Wait, that seems straightforward. So, part (I) is done.Part (II) says: If the distance from point P to the line y = x is √2 / 2, find the equation of circle P.Hmm, okay. So, point P is the center (a, b). The distance from (a, b) to the line y = x is given by the formula for the distance from a point to a line:Distance = |Ax + By + C| / √(A² + B²)For the line y = x, we can write it as x - y = 0. So, A = 1, B = -1, C = 0.Thus, the distance is |a - b| / √(1 + 1) = |a - b| / √2.According to the problem, this distance is √2 / 2. So,|a - b| / √2 = √2 / 2Multiply both sides by √2:|a - b| = (√2 / 2) * √2 = (2) / 2 = 1So, |a - b| = 1. Therefore, a - b = 1 or a - b = -1.So, now we have three equations:1. R² - b² = 22. R² - a² = 33. |a - b| = 1From equations 1 and 2, as before, we can subtract them to get b² - a² = 1.So, we have:b² - a² = 1and|a - b| = 1Let me write |a - b| = 1 as two cases:Case 1: a - b = 1Case 2: b - a = 1Let me handle each case separately.Case 1: a - b = 1 => a = b + 1Substitute a = b + 1 into b² - a² = 1:b² - (b + 1)² = 1Expand (b + 1)²: b² + 2b + 1So, b² - (b² + 2b + 1) = 1Simplify: b² - b² - 2b - 1 = 1Which is: -2b - 1 = 1So, -2b = 2 => b = -1Then, a = b + 1 = -1 + 1 = 0So, center is (0, -1)Now, find R. From equation 1: R² - b² = 2 => R² - (-1)² = 2 => R² - 1 = 2 => R² = 3 => R = √3So, equation of the circle is (x - 0)² + (y - (-1))² = 3 => x² + (y + 1)² = 3Case 2: b - a = 1 => b = a + 1Substitute b = a + 1 into b² - a² = 1:(a + 1)² - a² = 1Expand (a + 1)²: a² + 2a + 1So, (a² + 2a + 1) - a² = 1Simplify: 2a + 1 = 1So, 2a = 0 => a = 0Then, b = a + 1 = 0 + 1 = 1So, center is (0, 1)Find R: From equation 1: R² - b² = 2 => R² - (1)² = 2 => R² - 1 = 2 => R² = 3 => R = √3So, equation of the circle is (x - 0)² + (y - 1)² = 3 => x² + (y - 1)² = 3Therefore, there are two possible circles satisfying the conditions: one centered at (0, 1) and the other at (0, -1), both with radius √3.Wait, let me double-check if these satisfy the distance condition.For center (0, 1): distance to y = x is |0 - 1| / √2 = 1 / √2 = √2 / 2. Correct.For center (0, -1): distance is |0 - (-1)| / √2 = 1 / √2 = √2 / 2. Correct.So, both circles satisfy the distance condition.So, part (II) has two solutions.Wait, but in part (I), the trajectory is y² - x² = 1, which is a hyperbola. So, the centers lie on this hyperbola, and in part (II), we found specific points on this hyperbola that are at distance √2 / 2 from y = x.Yes, that makes sense.I think that's all. So, summarizing:(I) The trajectory equation is y² - x² = 1.(II) The equations of the circles are x² + (y - 1)² = 3 and x² + (y + 1)² = 3.**Final Answer**(I) The trajectory equation of the center of the circle ( P ) is (boxed{y^2 - x^2 = 1}).(II) The equations of circle ( P ) are (boxed{x^2 + (y - 1)^2 = 3}) and (boxed{x^2 + (y + 1)^2 = 3}).

💡Okay, so I need to figure out how much Katie spent in total at April's Discount Flowers. She bought different types of flowers with varying prices, and I need to calculate the total amount she spent. Let me break this down step by step to make sure I understand everything correctly.First, let's list out what Katie bought:- 5 roses at 6 each- 3 daisies at 4 each- 2 tulips at 5 eachAlright, so she bought three types of flowers: roses, daisies, and tulips. Each type has a different price per flower, and she bought a different quantity of each. To find the total amount she spent, I need to calculate how much she spent on each type of flower individually and then add those amounts together.Starting with the roses: she bought 5 roses, and each rose costs 6. So, to find the total cost for the roses, I need to multiply the number of roses by the price per rose. That would be 5 times 6. Let me write that out:5 roses * 6 per rose = 30Okay, so she spent 30 on roses. That seems straightforward.Next, let's move on to the daisies. She bought 3 daisies, and each daisy costs 4. Similar to the roses, I'll multiply the number of daisies by the price per daisy:3 daisies * 4 per daisy = 12So, she spent 12 on daisies. That's also clear.Now, onto the tulips. She bought 2 tulips, and each tulip costs 5. Again, I'll multiply the number of tulips by the price per tulip:2 tulips * 5 per tulip = 10Alright, she spent 10 on tulips.Now that I have the individual costs for each type of flower, I need to add them all together to find the total amount she spent. So, adding up the costs:30 (roses) + 12 (daisies) + 10 (tulips) = ?Let me do the addition step by step to avoid mistakes:First, add the cost of roses and daisies:30 + 12 = 42Now, add the cost of tulips to this sum:42 + 10 = 52So, when I add all the amounts together, the total comes out to 52.Wait a minute, let me double-check my calculations to make sure I didn't make any errors. It's easy to make a mistake with arithmetic, so better safe than sorry.Starting with the roses:5 roses * 6 = 30. That seems correct.Daisies:3 daisies * 4 = 12. Yep, that's right.Tulips:2 tulips * 5 = 10. That checks out too.Now, adding them up:30 + 12 = 4242 + 10 = 52Yes, that's consistent. So, Katie spent a total of 52 at April's Discount Flowers.I think I've covered all the steps and double-checked my work to ensure accuracy. It's always good to verify, especially when dealing with money and multiple calculations. This way, I can be confident that the total amount Katie spent is indeed 52.

💡First, I notice that each term in the expression (3a^{2} + 6a + 3) has a common factor of 3. I'll start by factoring out this greatest common factor.After factoring out the 3, the expression inside the parentheses becomes (a^{2} + 2a + 1). I recognize that this is a perfect square trinomial because it fits the form (a^{2} + 2ab + b^{2}), where (a = a) and (b = 1).Finally, I rewrite the perfect square trinomial as ((a + 1)^{2}) and include the factored out 3 to get the fully factorized form of the original expression.

💡First, I need to understand what constitutes a two-digit number. Two-digit numbers range from 10 to 99, inclusive.To find the total number of two-digit numbers, I'll subtract the smallest two-digit number from the largest one and then add one to account for both endpoints.So, 99 minus 10 equals 89. Adding one gives me 90.Therefore, there are 90 two-digit numbers.