💡Okay, so I have this trigonometry problem here: Given that sin(2α) equals negative sin(α), and α is between π/2 and π, I need to find the value of tan(α). Hmm, let's see. I remember that sin(2α) is a double angle formula, so maybe I can use that to rewrite the equation.Alright, the double angle formula for sine is sin(2α) = 2 sin(α) cos(α). So if I substitute that into the equation, I get 2 sin(α) cos(α) = -sin(α). Hmm, that looks like I can factor out sin(α) from both sides. Let me try that.So, factoring out sin(α), I get sin(α)(2 cos(α) + 1) = 0. Now, for this product to be zero, either sin(α) is zero or 2 cos(α) + 1 is zero. But wait, α is between π/2 and π, which is the second quadrant. In the second quadrant, sin(α) is positive, so sin(α) can't be zero here. That means the other factor must be zero. So, 2 cos(α) + 1 = 0.Let me solve for cos(α). Subtracting 1 from both sides gives 2 cos(α) = -1, and then dividing both sides by 2, I get cos(α) = -1/2. Okay, so cos(α) is -1/2. Now, I need to find α such that cos(α) is -1/2 and α is in the second quadrant.I remember that cos(α) = -1/2 corresponds to standard angles. Specifically, in the second quadrant, the angle whose cosine is -1/2 is 2π/3. Let me check that. Cos(2π/3) is indeed -1/2 because 2π/3 is 120 degrees, and cosine of 120 degrees is -1/2. So, α must be 2π/3.Now, I need to find tan(α). Since α is 2π/3, tan(2π/3) is tan(120 degrees). I know that tan(120 degrees) is negative because tangent is negative in the second quadrant. The reference angle for 120 degrees is 60 degrees, and tan(60 degrees) is √3. So, tan(120 degrees) is -√3.Therefore, tan(α) is -√3. Let me just recap to make sure I didn't make any mistakes. Starting from sin(2α) = -sin(α), I used the double angle formula, factored out sin(α), determined that cos(α) must be -1/2, found α as 2π/3, and then calculated tan(α) as -√3. It all seems to check out.I don't think I made any errors in the steps. I considered the quadrants correctly, used the right reference angles, and applied the trigonometric identities properly. So, I'm confident that the value of tan(α) is indeed -√3.**Final Answer**The value of tan alpha is boxed{-sqrt{3}}.

💡Okay, so I have this problem here: I need to find how many values of ( k ) make ( 12^{12} ) the least common multiple (LCM) of the numbers ( 6^6 ), ( 8^8 ), and ( k ). Additionally, ( k ) has to be of the form ( 2^a 3^b 5^c ). Hmm, let me try to break this down step by step.First, I remember that the LCM of several numbers is the smallest number that is a multiple of each of them. So, if I take the LCM of ( 6^6 ), ( 8^8 ), and ( k ), it should equal ( 12^{12} ). To find ( k ), I probably need to express all these numbers in terms of their prime factors because LCM is easier to handle that way.Let me start by factoring each number:- ( 6^6 ): Since 6 is ( 2 times 3 ), raising it to the 6th power gives ( (2 times 3)^6 = 2^6 times 3^6 ).- ( 8^8 ): 8 is ( 2^3 ), so ( (2^3)^8 = 2^{24} ).- ( 12^{12} ): 12 is ( 2^2 times 3 ), so ( (2^2 times 3)^{12} = 2^{24} times 3^{12} ).Okay, so now I have:- ( 6^6 = 2^6 times 3^6 )- ( 8^8 = 2^{24} )- ( 12^{12} = 2^{24} times 3^{12} )Now, ( k ) is given as ( 2^a 3^b 5^c ). So, ( k ) can have factors of 2, 3, and 5, but no other primes. The next step is to find the LCM of ( 6^6 ), ( 8^8 ), and ( k ). The LCM of multiple numbers takes the highest power of each prime present in the numbers. So, let me compute the LCM step by step.First, let's find the LCM of ( 6^6 ) and ( 8^8 ):- For prime 2: The highest power is ( 2^{24} ) from ( 8^8 ).- For prime 3: The highest power is ( 3^6 ) from ( 6^6 ).So, ( text{LCM}(6^6, 8^8) = 2^{24} times 3^6 ).Now, we need to find the LCM of this result with ( k ), which is ( 2^a 3^b 5^c ). So, let's compute ( text{LCM}(2^{24} times 3^6, 2^a 3^b 5^c) ).Again, taking the highest powers for each prime:- For prime 2: The highest power is ( max(24, a) ).- For prime 3: The highest power is ( max(6, b) ).- For prime 5: The highest power is ( max(0, c) ) since the previous LCM didn't have any factor of 5.We want this LCM to be equal to ( 12^{12} = 2^{24} times 3^{12} ). So, setting up the equation:[ 2^{max(24, a)} times 3^{max(6, b)} times 5^{max(0, c)} = 2^{24} times 3^{12} ]Now, let's analyze each prime factor:1. **For prime 2:** - The exponent on the left side is ( max(24, a) ). - On the right side, it's 24. - Therefore, ( max(24, a) = 24 ). This implies that ( a leq 24 ). Because if ( a ) were greater than 24, the max would be ( a ), which would exceed 24, but we need it to be exactly 24.2. **For prime 3:** - The exponent on the left side is ( max(6, b) ). - On the right side, it's 12. - So, ( max(6, b) = 12 ). This means that ( b ) must be 12. Because if ( b ) were less than 12, the max would be 6 or something else, but we need it to be 12. If ( b ) is exactly 12, then the max is 12, which matches the right side.3. **For prime 5:** - The exponent on the left side is ( max(0, c) ). - On the right side, there is no factor of 5, so the exponent is 0. - Therefore, ( max(0, c) = 0 ). This implies that ( c = 0 ). Because if ( c ) were greater than 0, the max would be ( c ), which would introduce a factor of 5, but we need the exponent to remain 0.So, putting it all together, we have:- ( a leq 24 )- ( b = 12 )- ( c = 0 )Now, ( a ) can be any integer from 0 up to 24, inclusive. That gives us 25 possible values for ( a ) (since 0 is included). ( b ) is fixed at 12, so there's only 1 choice for ( b ). Similarly, ( c ) is fixed at 0, so only 1 choice for ( c ).Therefore, the total number of possible values for ( k ) is the product of the number of choices for each exponent:[ 25 times 1 times 1 = 25 ]So, there are 25 possible values of ( k ) that satisfy the given conditions.Wait, let me double-check to make sure I didn't miss anything. - For ( a ), since it's ( 2^a ), and ( a ) can be from 0 to 24, inclusive, that's 25 values. That seems right.- For ( b ), it has to be exactly 12 because if it were less, the LCM wouldn't reach 12, and if it were more, the LCM would exceed 12, which isn't allowed. So, only 1 value.- For ( c ), it has to be 0 because any positive exponent would introduce a factor of 5, which isn't present in ( 12^{12} ). So, only 1 value.Multiplying these together gives 25, which seems correct. I don't think I missed any constraints or possibilities here. So, yeah, 25 is the answer.**Final Answer**The number of possible values of ( k ) is boxed{25}.

💡First, I recognize that the problem involves a ratio of 150 to 1 between two numbers. The certain number is given as 300, and I need to find the other number.I set up the proportion based on the given ratio:300 : Other number = 150 : 1To solve for the other number, I cross-multiply:300 * 1 = 150 * Other numberThis simplifies to:300 = 150 * Other numberFinally, I divide both sides by 150 to isolate the other number:Other number = 300 / 150Other number = 2Therefore, the other number is 2.

💡Okay, so I have this problem about a real polynomial P with some conditions, and I need to show that P(n) is positive for all natural numbers n. Let me try to unpack this step by step.First, the problem states that P(0) > 0, which is straightforward. Then, P(1) > P(0), so the value of the polynomial at 1 is greater than at 0. Next, P(2) > 2P(1) - P(0). Hmm, that looks like a condition on the second difference of the polynomial. And finally, for all natural numbers n, P(n+3) > 3P(n+2) - 3P(n+1) + P(n). That seems like a condition on the third difference.I remember that for polynomials, the k-th difference is related to the k-th derivative in calculus, but in discrete terms. So maybe I can use some kind of discrete calculus here.Let me define the first difference as Δ₁(X) = P(X+1) - P(X). Then the second difference would be Δ₂(X) = Δ₁(X+1) - Δ₁(X) = P(X+2) - 2P(X+1) + P(X). Similarly, the third difference Δ₃(X) = Δ₂(X+1) - Δ₂(X) = P(X+3) - 3P(X+2) + 3P(X+1) - P(X).Given the condition P(n+3) > 3P(n+2) - 3P(n+1) + P(n), that means Δ₃(n) > 0 for all n in natural numbers. So the third difference is always positive.Now, if the third difference is positive, what does that say about the second difference? Well, if Δ₃(n) > 0, that means Δ₂(n+1) - Δ₂(n) > 0, so Δ₂(n+1) > Δ₂(n). Therefore, the second difference is strictly increasing.We also have a condition on the second difference at n=0: P(2) > 2P(1) - P(0), which is exactly Δ₂(0) > 0. Since Δ₂(n) is strictly increasing and starts positive, it must remain positive for all n. So Δ₂(n) > 0 for all n.Now, moving up to the first difference. Since Δ₂(n) > 0, that means Δ₁(n+1) - Δ₁(n) > 0, so Δ₁(n+1) > Δ₁(n). Therefore, the first difference is also strictly increasing.We have Δ₁(0) = P(1) - P(0) > 0, as given. Since Δ₁(n) is strictly increasing and starts positive, it must remain positive for all n. So Δ₁(n) > 0 for all n.Finally, since Δ₁(n) > 0, that means P(n+1) - P(n) > 0, so P(n+1) > P(n). Starting from P(0) > 0 and knowing that P(n) is strictly increasing, it follows that P(n) > 0 for all n in natural numbers.Wait, let me double-check. If the third difference is positive, the second difference increases, which makes the first difference increase, and since the first difference is positive, the polynomial itself increases. So yes, starting from a positive value and increasing, P(n) stays positive.I think that covers all the steps. I should make sure there are no gaps in the logic. The key was recognizing the conditions on the differences and translating them into properties of the polynomial's behavior. Since all the differences are positive and increasing, the polynomial itself must be positive and increasing for all natural numbers n.

💡Okay, so I have this geometry problem here, and I need to prove that point D is the midpoint of segment EF. Let me try to visualize the problem first.We have a circle with center O, and a point P outside the circle. From P, there are two tangent lines PA and PB, touching the circle at points A and B respectively. So, PA and PB are both tangents from P to the circle. That means PA equals PB because the lengths of tangents from a common external point to a circle are equal. That's a property I remember.Now, there's another point C on the circle. A tangent line at C intersects PA and PB at points E and F respectively. So, the tangent at C meets PA at E and PB at F. That gives us points E and F on PA and PB.Then, line OC intersects AB at point L. So, connecting the center O to point C, and this line meets AB at L. Then, connecting L to P, this line LP intersects EF at point D. We need to show that D is the midpoint of EF.Alright, so let me try to draw this in my mind. We have the circle, point P outside, tangents PA and PB, tangent at C meets PA at E and PB at F, OC meets AB at L, and LP meets EF at D. We need to prove D is the midpoint of EF.First, maybe I can recall some properties of harmonic division, projective geometry, or maybe use similar triangles. Alternatively, coordinate geometry might be an approach, but that might get messy. Let me think about synthetic geometry first.Since PA and PB are tangents, OA is perpendicular to PA, and OB is perpendicular to PB. So, OA ⊥ PA and OB ⊥ PB. That gives us right angles at A and B.Also, since E and F lie on the tangent at C, EC and FC are tangents from E and F to the circle. Wait, no, actually, the tangent at C passes through E and F, so EC and FC are parts of the same tangent line. So, EC = FC? Wait, no, because E and F are on PA and PB, so EC and FC are not necessarily equal. Hmm.Wait, but since E and F are on the tangent at C, the lengths from E to C and F to C are equal? No, that's not necessarily the case because E and F are on different lines PA and PB. Hmm, maybe not.Alternatively, maybe I can use the power of a point. The power of point E with respect to the circle is equal to EC², since EC is tangent to the circle. Similarly, the power of E is also equal to EA * EP, because E lies on PA, which is a secant. So, EC² = EA * EP. Similarly, for point F, FC² = FB * FP.But since EC and FC are the same tangent from E and F to the circle, wait, no, E and F are on the same tangent line at C, so EC and FC are the same line segment, so EC = FC. Wait, no, E and F are on the tangent line at C, so EC and FC are parts of the same tangent line, but E and F are different points on that tangent. So, EC and FC are not necessarily equal. Hmm, maybe that's not the right approach.Alternatively, maybe I can consider triangle PAB. Since PA and PB are tangents, triangle PAB is isoceles with PA = PB. Then, AB is the base of this isoceles triangle. Point L is the intersection of OC and AB. Maybe I can find some properties about L.Also, since OC is a radius, and C is a point on the circle, OC is just a radius. So, OC is equal to OA and OB.Wait, maybe I can use harmonic conjugates or something like that. Since PA and PB are tangents, and E and F are points where another tangent intersects PA and PB, maybe there's a harmonic bundle here.Alternatively, maybe using Ceva's theorem or Menelaus' theorem. Let me think about Ceva's theorem. If I can find concurrent lines, maybe that could help.Wait, let's consider triangle PAB. Points E and F are on PA and PB respectively, and line EF intersects AB at some point. Wait, but in our case, line EF is intersected by LP at D. Maybe I can apply Menelaus' theorem to triangle PAB with transversal EF.Wait, Menelaus' theorem states that for a triangle, if a line crosses the three sides (or their extensions), the product of the segment ratios is equal to 1. So, for triangle PAB, if line EF intersects PA at E, PB at F, and AB at some point, say M, then (PE/EA) * (AF/FB) * (BM/MP) = 1.But in our case, EF is intersected by LP at D. So, maybe I can relate this to the ratios.Alternatively, maybe using projective geometry, considering the pole-polar relationships. Since PA and PB are tangents, their poles are A and B. The polar of P is line AB.Similarly, the polar of point C is the tangent line at C, which is EF. So, the polar of C is EF.Now, point L is the intersection of OC and AB. Since AB is the polar of P, and OC is a line from the center to C, which is on the circle. Maybe there's a relationship here.Wait, in projective geometry, the pole of AB is P, and the pole of EF is C. So, maybe there's a reciprocal relationship.Alternatively, maybe using La Hire's theorem, which states that if a point lies on the polar of another point, then the latter lies on the polar of the former. So, since L is on AB, which is the polar of P, then P lies on the polar of L. Similarly, since L is on OC, which is a line from the center, maybe we can find something about the polar of L.Wait, the polar of L would pass through P, because L is on AB, the polar of P. So, polar(L) passes through P.Also, since L is on OC, which is a radius, maybe we can find the equation of polar(L). But I'm not sure if this is the right path.Alternatively, maybe using harmonic division. Since PA and PB are tangents, and E and F are points where another tangent intersects PA and PB, maybe the division is harmonic.Wait, let's consider the complete quadrilateral formed by PA, PB, EF, and AB. The intersection points are P, A, B, E, F, L, D. Maybe using properties of complete quadrilaterals.Alternatively, maybe using inversion. If I invert the figure with respect to the circle, point P would invert to some point, tangents would invert to circles, but I'm not sure if that would simplify things.Wait, maybe I can use similar triangles. Let me see.Since OA is perpendicular to PA, and OC is a radius, maybe triangle OAP is similar to some other triangle.Alternatively, since OC intersects AB at L, maybe I can find some ratio involving OL and LC.Wait, let me try to find some similar triangles.Looking at triangles OAP and OCP. Wait, OA = OC because they are radii, and OP is common. But angle OAP is 90 degrees, and angle OCP is also 90 degrees because OC is perpendicular to the tangent at C. So, triangles OAP and OCP are both right triangles with a common hypotenuse OP. Therefore, they are congruent. So, OA = OC, OP is common, and right angles, so by hypotenuse-leg congruence, triangles OAP and OCP are congruent.Therefore, angle OPA equals angle OPC. So, OP bisects angle APC. Hmm, interesting.Wait, but angle APC is the angle between PA and PC. Since PA is tangent at A, and PC is another line from P to C on the circle. So, OP bisects angle APC.Similarly, OP bisects angle BPC as well, since triangle OPB is congruent to triangle OPC.Wait, maybe that's useful.Alternatively, maybe I can use the fact that since OP bisects angle APC, and E is on PA, F is on PB, maybe there's some ratio that can be established.Alternatively, maybe using the harmonic conjugate. Since PA and PB are tangents, and E and F are on the tangent at C, maybe the cross ratio is harmonic.Wait, let me try to think about the cross ratio. If I consider the pencil from P, with lines PA, PB, PE, PF, maybe there's a harmonic bundle.Alternatively, maybe using the fact that EF is the polar of C, and L is on AB, which is the polar of P. So, maybe there's a reciprocal relationship.Wait, if I consider the polar of L, since L is on AB, which is the polar of P, then P lies on the polar of L. Also, since L is on OC, which is the line from the center to C, the polar of L would pass through the inverse point of L with respect to the circle.Wait, maybe this is getting too abstract.Alternatively, maybe I can use coordinates. Let me try setting up a coordinate system.Let me place the circle with center at the origin (0,0), and let the circle have radius r. Let me place point P outside the circle, say at (0, k), where k > r. Then, the tangents from P to the circle will touch the circle at points A and B.The equation of the circle is x² + y² = r². The equation of the tangent from P(0,k) to the circle can be found using the formula for tangent from an external point.The equation of tangent from (0,k) is xx1 + yy1 = r², where (x1,y1) is the point of contact. But since P is (0,k), the tangent will have the form y = mx + c, and since it passes through (0,k), c = k. So, the tangent equation is y = mx + k.The condition for this line to be tangent to the circle x² + y² = r² is that the distance from the center (0,0) to the line is equal to r. The distance from (0,0) to y = mx + k is |0 - 0 + k| / sqrt(m² + 1) = |k| / sqrt(m² + 1) = r.So, |k| / sqrt(m² + 1) = r => sqrt(m² + 1) = |k| / r => m² + 1 = k² / r² => m² = (k² / r²) - 1.So, m = ±sqrt((k² - r²)/r²) = ±sqrt(k² - r²)/r.Therefore, the equations of the tangents are y = [sqrt(k² - r²)/r]x + k and y = [-sqrt(k² - r²)/r]x + k.So, points A and B are the points of contact of these tangents with the circle.Let me find the coordinates of A and B.For the tangent y = [sqrt(k² - r²)/r]x + k, solving with the circle x² + y² = r².Substitute y into the circle equation:x² + ([sqrt(k² - r²)/r]x + k)² = r².Expanding this:x² + ( (k² - r²)/r² x² + 2*sqrt(k² - r²)/r * k x + k² ) = r².Combine like terms:x² + (k² - r²)/r² x² + 2k sqrt(k² - r²)/r x + k² - r² = 0.Let me factor out x²:[1 + (k² - r²)/r²] x² + 2k sqrt(k² - r²)/r x + (k² - r²) = 0.Simplify the coefficient of x²:1 + (k² - r²)/r² = (r² + k² - r²)/r² = k² / r².So, the equation becomes:(k² / r²) x² + 2k sqrt(k² - r²)/r x + (k² - r²) = 0.Multiply both sides by r² to eliminate denominators:k² x² + 2k r sqrt(k² - r²) x + (k² - r²) r² = 0.This is a quadratic in x. Since the line is tangent, it should have a single solution, so discriminant is zero.But actually, we can find the point of contact. The x-coordinate of the point of contact can be found using the formula for the point of contact of a tangent from an external point.Alternatively, since we know the tangent touches the circle at A, and the tangent line is y = [sqrt(k² - r²)/r]x + k, the point A can be found by solving the system.But maybe it's easier to parameterize point C on the circle and find the tangent at C, then find E and F.Let me parameterize point C on the circle as (r cos θ, r sin θ). Then, the tangent at C is x cos θ + y sin θ = r.This tangent intersects PA and PB at points E and F.First, let me find the equations of PA and PB.From earlier, PA is y = [sqrt(k² - r²)/r]x + k, and PB is y = [-sqrt(k² - r²)/r]x + k.So, to find E, which is the intersection of tangent at C and PA.So, solving:x cos θ + y sin θ = r,andy = [sqrt(k² - r²)/r]x + k.Substitute y from the second equation into the first:x cos θ + ([sqrt(k² - r²)/r]x + k) sin θ = r.Simplify:x cos θ + [sqrt(k² - r²)/r]x sin θ + k sin θ = r.Factor x:x [cos θ + (sqrt(k² - r²)/r) sin θ] + k sin θ = r.Solve for x:x = (r - k sin θ) / [cos θ + (sqrt(k² - r²)/r) sin θ].Similarly, the y-coordinate of E is:y = [sqrt(k² - r²)/r]x + k.Similarly, for point F, which is the intersection of tangent at C and PB.PB has equation y = [-sqrt(k² - r²)/r]x + k.So, solving:x cos θ + y sin θ = r,andy = [-sqrt(k² - r²)/r]x + k.Substitute y into the first equation:x cos θ + [(-sqrt(k² - r²)/r)x + k] sin θ = r.Simplify:x cos θ - [sqrt(k² - r²)/r]x sin θ + k sin θ = r.Factor x:x [cos θ - (sqrt(k² - r²)/r) sin θ] + k sin θ = r.Solve for x:x = (r - k sin θ) / [cos θ - (sqrt(k² - r²)/r) sin θ].And the y-coordinate of F is:y = [-sqrt(k² - r²)/r]x + k.Now, we have coordinates for E and F in terms of θ, k, and r.Now, we need to find point L, which is the intersection of OC and AB.First, let's find the equation of AB.Points A and B are the points of contact of the tangents from P(0,k). From earlier, we can find their coordinates.Alternatively, since we have the equations of PA and PB, we can find points A and B as the points where these tangents touch the circle.Alternatively, since we have the equations of PA and PB, we can find their intersection with the circle.But maybe it's easier to find AB as the polar of P.Since P is (0,k), the polar of P is the line AB. The equation of the polar of P with respect to the circle x² + y² = r² is given by:0*x + k*y = r² => y = r² / k.So, AB is the line y = r² / k.Therefore, the equation of AB is y = r² / k.Now, line OC is the line from the center O(0,0) to point C(r cos θ, r sin θ). So, the parametric equation of OC is x = r cos θ * t, y = r sin θ * t, where t is a parameter.We need to find the intersection point L of OC and AB.AB is y = r² / k.So, set y = r sin θ * t = r² / k.Solve for t:t = (r² / k) / (r sin θ) = r / (k sin θ).Therefore, the coordinates of L are:x = r cos θ * t = r cos θ * (r / (k sin θ)) = (r² cos θ) / (k sin θ),y = r sin θ * t = r sin θ * (r / (k sin θ)) = r² / k.So, L is at ((r² cos θ)/(k sin θ), r² / k).Now, we need to find the equation of line LP, which connects L and P(0,k).The coordinates of L are ((r² cos θ)/(k sin θ), r² / k), and P is (0,k).So, the slope of LP is:m = (k - r² / k) / (0 - (r² cos θ)/(k sin θ)) = ( (k² - r²)/k ) / ( - r² cos θ / (k sin θ) ) = ( (k² - r²)/k ) * ( - k sin θ / (r² cos θ) ) = - (k² - r²) sin θ / (r² cos θ).Therefore, the equation of LP is:y - k = m (x - 0) = - (k² - r²) sin θ / (r² cos θ) x.So, y = - (k² - r²) sin θ / (r² cos θ) x + k.Now, we need to find the intersection point D of LP and EF.First, let's find the equation of EF.Points E and F are on PA and PB respectively, and their coordinates are known in terms of θ, k, and r.But this might get complicated. Alternatively, since EF is the tangent at C, which has equation x cos θ + y sin θ = r.Wait, no, EF is the tangent at C, so its equation is x cos θ + y sin θ = r.Wait, but earlier, we found E and F as the intersections of this tangent with PA and PB. So, EF is the tangent line at C, which is x cos θ + y sin θ = r.Therefore, the equation of EF is x cos θ + y sin θ = r.So, to find point D, we need to solve the system:y = - (k² - r²) sin θ / (r² cos θ) x + k,andx cos θ + y sin θ = r.Substitute y from the first equation into the second equation:x cos θ + [ - (k² - r²) sin θ / (r² cos θ) x + k ] sin θ = r.Simplify:x cos θ - (k² - r²) sin² θ / (r² cos θ) x + k sin θ = r.Factor x:x [ cos θ - (k² - r²) sin² θ / (r² cos θ) ] + k sin θ = r.Let me combine the terms:x [ (r² cos² θ - (k² - r²) sin² θ ) / (r² cos θ) ] + k sin θ = r.So,x [ (r² cos² θ - k² sin² θ + r² sin² θ ) / (r² cos θ) ] + k sin θ = r.Simplify the numerator:r² cos² θ + r² sin² θ - k² sin² θ = r² (cos² θ + sin² θ) - k² sin² θ = r² - k² sin² θ.So, the equation becomes:x [ (r² - k² sin² θ ) / (r² cos θ) ] + k sin θ = r.Solve for x:x = [ r - k sin θ ] / [ (r² - k² sin² θ ) / (r² cos θ) ] = [ (r - k sin θ ) * r² cos θ ] / (r² - k² sin² θ ).Simplify numerator and denominator:Numerator: r² cos θ (r - k sin θ )Denominator: r² - k² sin² θ = (r - k sin θ)(r + k sin θ )So, x = [ r² cos θ (r - k sin θ ) ] / [ (r - k sin θ)(r + k sin θ ) ] = [ r² cos θ ] / (r + k sin θ ).Therefore, x = r² cos θ / (r + k sin θ ).Now, substitute x back into the equation of LP to find y:y = - (k² - r²) sin θ / (r² cos θ) * [ r² cos θ / (r + k sin θ ) ] + k.Simplify:y = - (k² - r²) sin θ / (r + k sin θ ) + k.Combine terms:y = [ - (k² - r²) sin θ + k (r + k sin θ ) ] / (r + k sin θ )Expand numerator:- k² sin θ + r² sin θ + k r + k² sin θ = r² sin θ + k r.So, y = (r² sin θ + k r ) / (r + k sin θ ) = r (r sin θ + k ) / (r + k sin θ ) = r.Wait, that's interesting. So, y = r.Wait, but y = r is a horizontal line. But point D is supposed to be on EF, which is x cos θ + y sin θ = r.If y = r, then substituting into EF's equation:x cos θ + r sin θ = r => x cos θ = r (1 - sin θ ) => x = r (1 - sin θ ) / cos θ.But earlier, we found x = r² cos θ / (r + k sin θ ).Wait, but if y = r, then x must satisfy both equations.Wait, maybe I made a mistake in the calculation.Let me check the calculation for y.We had:y = - (k² - r²) sin θ / (r² cos θ) * x + k,and x = r² cos θ / (r + k sin θ ).So,y = - (k² - r²) sin θ / (r² cos θ) * [ r² cos θ / (r + k sin θ ) ] + kSimplify:The r² cos θ cancels out:y = - (k² - r²) sin θ / (r + k sin θ ) + kNow, combine terms:y = [ - (k² - r²) sin θ + k (r + k sin θ ) ] / (r + k sin θ )Expand numerator:- k² sin θ + r² sin θ + k r + k² sin θ = r² sin θ + k rSo, numerator is r² sin θ + k r = r (r sin θ + k )Therefore,y = r (r sin θ + k ) / (r + k sin θ ) = r.So, y = r.Therefore, point D has coordinates ( r² cos θ / (r + k sin θ ), r ).Now, we need to check if D is the midpoint of EF.Points E and F are on PA and PB respectively, and their coordinates are known.From earlier, E has coordinates:x_E = (r - k sin θ ) / [cos θ + (sqrt(k² - r²)/r) sin θ ],y_E = [sqrt(k² - r²)/r] x_E + k.Similarly, F has coordinates:x_F = (r - k sin θ ) / [cos θ - (sqrt(k² - r²)/r) sin θ ],y_F = [-sqrt(k² - r²)/r] x_F + k.This seems complicated, but maybe we can find the midpoint of EF and see if it equals D.Let me denote sqrt(k² - r²) as m for simplicity.So, m = sqrt(k² - r²).Then,x_E = (r - k sin θ ) / [cos θ + (m/r) sin θ ],y_E = (m/r) x_E + k,x_F = (r - k sin θ ) / [cos θ - (m/r) sin θ ],y_F = (-m/r) x_F + k.Now, the midpoint M of EF would have coordinates:x_M = (x_E + x_F)/2,y_M = (y_E + y_F)/2.Let me compute x_M:x_M = [ (r - k sin θ ) / (cos θ + (m/r) sin θ ) + (r - k sin θ ) / (cos θ - (m/r) sin θ ) ] / 2.Factor out (r - k sin θ ):x_M = (r - k sin θ ) [ 1 / (cos θ + (m/r) sin θ ) + 1 / (cos θ - (m/r) sin θ ) ] / 2.Combine the fractions:= (r - k sin θ ) [ (cos θ - (m/r) sin θ + cos θ + (m/r) sin θ ) / ( (cos θ + (m/r) sin θ )(cos θ - (m/r) sin θ ) ) ] / 2.Simplify numerator:= (r - k sin θ ) [ (2 cos θ ) / (cos² θ - (m² / r² ) sin² θ ) ] / 2.So,x_M = (r - k sin θ ) * 2 cos θ / [ 2 (cos² θ - (m² / r² ) sin² θ ) ] = (r - k sin θ ) cos θ / (cos² θ - (m² / r² ) sin² θ ).Now, let's compute the denominator:cos² θ - (m² / r² ) sin² θ = cos² θ - ( (k² - r²)/r² ) sin² θ = (r² cos² θ - (k² - r² ) sin² θ ) / r².So,x_M = (r - k sin θ ) cos θ / [ (r² cos² θ - (k² - r² ) sin² θ ) / r² ] = (r - k sin θ ) cos θ * r² / (r² cos² θ - k² sin² θ + r² sin² θ ).Simplify denominator:r² cos² θ + r² sin² θ - k² sin² θ = r² (cos² θ + sin² θ ) - k² sin² θ = r² - k² sin² θ.So,x_M = (r - k sin θ ) cos θ * r² / (r² - k² sin² θ ).Factor numerator:= r² cos θ (r - k sin θ ) / (r² - k² sin² θ ).But notice that r² - k² sin² θ = (r - k sin θ )(r + k sin θ ).So,x_M = r² cos θ (r - k sin θ ) / [ (r - k sin θ )(r + k sin θ ) ] = r² cos θ / (r + k sin θ ).Which is exactly the x-coordinate of D.Similarly, let's compute y_M:y_M = (y_E + y_F ) / 2.From earlier,y_E = (m/r) x_E + k,y_F = (-m/r) x_F + k.So,y_M = [ (m/r) x_E + k + (-m/r) x_F + k ] / 2 = [ (m/r)(x_E - x_F ) + 2k ] / 2.Compute x_E - x_F:x_E - x_F = (r - k sin θ ) [ 1 / (cos θ + (m/r) sin θ ) - 1 / (cos θ - (m/r) sin θ ) ].= (r - k sin θ ) [ (cos θ - (m/r) sin θ - cos θ - (m/r) sin θ ) / ( (cos θ + (m/r) sin θ )(cos θ - (m/r) sin θ ) ) ]= (r - k sin θ ) [ ( - 2 (m/r ) sin θ ) / (cos² θ - (m² / r² ) sin² θ ) ]= -2 (m/r ) sin θ (r - k sin θ ) / (cos² θ - (m² / r² ) sin² θ ).So,y_M = [ (m/r ) * ( -2 (m/r ) sin θ (r - k sin θ ) / (cos² θ - (m² / r² ) sin² θ ) ) + 2k ] / 2.Simplify:= [ -2 (m² / r² ) sin θ (r - k sin θ ) / (cos² θ - (m² / r² ) sin² θ ) + 2k ] / 2.Factor out 2:= [ - (m² / r² ) sin θ (r - k sin θ ) / (cos² θ - (m² / r² ) sin² θ ) + k ].Now, substitute m² = k² - r²:= [ - ( (k² - r² ) / r² ) sin θ (r - k sin θ ) / (cos² θ - ( (k² - r² ) / r² ) sin² θ ) + k ].Simplify denominator:cos² θ - ( (k² - r² ) / r² ) sin² θ = (r² cos² θ - (k² - r² ) sin² θ ) / r² = (r² - k² sin² θ ) / r².So,= [ - ( (k² - r² ) / r² ) sin θ (r - k sin θ ) / ( (r² - k² sin² θ ) / r² ) + k ].Simplify fractions:= [ - ( (k² - r² ) sin θ (r - k sin θ ) ) / (r² - k² sin² θ ) + k ].Factor numerator:= [ - ( (k² - r² ) sin θ (r - k sin θ ) ) / ( (r - k sin θ )(r + k sin θ ) ) + k ].Cancel (r - k sin θ ):= [ - ( (k² - r² ) sin θ ) / (r + k sin θ ) + k ].Now, note that (k² - r² ) = (k - r)(k + r ), but maybe we can write it as -(r² - k² ).Wait, let's write it as:= [ - ( - (r² - k² ) sin θ ) / (r + k sin θ ) + k ] = [ (r² - k² ) sin θ / (r + k sin θ ) + k ].Now, combine terms:= [ (r² - k² ) sin θ + k (r + k sin θ ) ] / (r + k sin θ ).Expand numerator:= r² sin θ - k² sin θ + k r + k² sin θ = r² sin θ + k r.So,y_M = (r² sin θ + k r ) / (r + k sin θ ) = r (r sin θ + k ) / (r + k sin θ ) = r.Therefore, y_M = r.So, the midpoint M of EF has coordinates ( r² cos θ / (r + k sin θ ), r ), which is exactly the same as point D.Therefore, D is the midpoint of EF.**Final Answer**boxed{D} text{ is the midpoint of } EF.

💡Okay, so I have this problem here about cylindrical coordinates. It says, for a positive constant ( d ), in cylindrical coordinates ( (r, theta, z) ), find the shape described by the equation ( r = d ). The options are Line, Circle, Plane, Sphere, Cylinder, Cone. I need to figure out which one it is.First, I remember that cylindrical coordinates are kind of like polar coordinates but extended into three dimensions. In polar coordinates, ( r ) is the distance from the origin, and ( theta ) is the angle. So in cylindrical coordinates, ( r ) is still the radial distance in the ( xy )-plane, ( theta ) is the angle around the ( z )-axis, and ( z ) is the height.So the equation ( r = d ) means that every point in this set has a radial distance of ( d ) from the ( z )-axis. That sounds like a cylinder because a cylinder has a constant radius around an axis. But wait, let me think more carefully.In polar coordinates, if I have ( r = d ), that's a circle with radius ( d ) in the ( xy )-plane. But here, in cylindrical coordinates, ( z ) can be anything, right? So it's not just a single circle; it's a circle that extends infinitely in both the positive and negative ( z )-directions. So that would make it a cylinder.Let me visualize this. Imagine the ( z )-axis going up and down. If I fix ( r = d ), I'm creating a circle in every horizontal plane (like the ( xy )-plane, ( xz )-plane, etc.) at every height ( z ). So connecting all these circles along the ( z )-axis gives me a cylinder.Is there any chance it could be something else? Let me check the other options. A line would mean all points lie along a straight path, but ( r = d ) isn't a line because it's a set of points at a fixed distance from the ( z )-axis, which forms a circular path. A circle is just a single loop, but since ( z ) can vary, it's not just one circle. A plane would be a flat, two-dimensional surface extending infinitely, but ( r = d ) is a curved surface, not flat. A sphere requires all three coordinates to be related in a way that forms a ball shape, which isn't the case here. A cone would have ( r ) proportional to ( z ), but here ( r ) is fixed, so it's not a cone.Therefore, the shape described by ( r = d ) in cylindrical coordinates is a cylinder.