answer：First, I need to determine the total number of three-digit integers less than 600. These integers range from 100 to 599, inclusive. To find the total count, I subtract the smallest number from the largest and add one:599 - 100 + 1 = 500Next, I'll calculate the number of three-digit integers where all digits are unique. The first digit (hundreds place) can be 1, 2, 3, 4, or 5, giving 5 options. The second digit (tens place) can be any of the remaining 9 digits (0-9 excluding the hundreds digit), and the third digit (units place) can be any of the remaining 8 digits (excluding the hundreds and tens digits):5 (hundreds place) × 9 (tens place) × 8 (units place) = 360Finally, to find the number of integers with at least two digits the same, I subtract the number of integers with all unique digits from the total number of integers:500 - 360 = 140Therefore, there are 140 three-digit integers less than 600 that have at least two digits that are the same.

answer：Alright, so I need to figure out which of these vector parameterizations correctly represent the line y = x - 1. Hmm, okay. Let me start by recalling what a vector parameterization of a line looks like. I think it's something like this:[begin{pmatrix} x y end{pmatrix} = mathbf{a} + t mathbf{d}]Where (mathbf{a}) is a point on the line, and (mathbf{d}) is the direction vector of the line. So, for the line y = x - 1, I need to make sure that the point (mathbf{a}) lies on the line and that the direction vector (mathbf{d}) has a slope that matches the line's slope.First, let's find the slope of the line y = x - 1. The equation is already in slope-intercept form, y = mx + b, where m is the slope. So, the slope m is 1. That means the direction vector should have a rise over run of 1, which translates to a direction vector like (begin{pmatrix} 1 1 end{pmatrix}) or any scalar multiple of that.Now, let's go through each option one by one.**Option A:**[begin{pmatrix} x y end{pmatrix} = begin{pmatrix} -1 -2 end{pmatrix} + t begin{pmatrix} 2 2 end{pmatrix}]Okay, so the point here is (-1, -2). Let me check if that lies on the line y = x - 1. Plugging in x = -1, y should be -1 - 1 = -2. So, yes, (-1, -2) is on the line. Good.Now, the direction vector is (begin{pmatrix} 2 2 end{pmatrix}). Let's see if this has a slope of 1. The slope is rise over run, which is 2/2 = 1. Perfect, that matches the slope of the line. So, Option A seems correct. Wait, but I remember sometimes direction vectors can be scaled, but as long as the slope is the same, it's fine. So, I think A is correct.**Option B:**[begin{pmatrix} x y end{pmatrix} = begin{pmatrix} 1 0 end{pmatrix} + t begin{pmatrix} 1 1 end{pmatrix}]Point is (1, 0). Let's check if it's on the line. Plugging x = 1 into y = x - 1, we get y = 0. So, yes, (1, 0) is on the line.Direction vector is (begin{pmatrix} 1 1 end{pmatrix}), which has a slope of 1/1 = 1. That's exactly the slope we need. So, Option B is definitely correct.**Option C:**[begin{pmatrix} x y end{pmatrix} = begin{pmatrix} 0 -1 end{pmatrix} + t begin{pmatrix} -1 -1 end{pmatrix}]Point is (0, -1). Checking if it's on the line: y = 0 - 1 = -1. Yes, it is.Direction vector is (begin{pmatrix} -1 -1 end{pmatrix}). Slope is (-1)/(-1) = 1. That's correct. So, Option C is also correct.**Option D:**[begin{pmatrix} x y end{pmatrix} = begin{pmatrix} 3 2 end{pmatrix} + t begin{pmatrix} 3 4 end{pmatrix}]Point is (3, 2). Let's see: y = 3 - 1 = 2. So, yes, it's on the line.Direction vector is (begin{pmatrix} 3 4 end{pmatrix}). Slope is 4/3, which is approximately 1.333. That's not equal to 1, so this direction vector doesn't match the slope of the line. Therefore, Option D is incorrect.**Option E:**[begin{pmatrix} x y end{pmatrix} = begin{pmatrix} 2 1 end{pmatrix} + t begin{pmatrix} 0.5 0.5 end{pmatrix}]Point is (2, 1). Checking: y = 2 - 1 = 1. Correct.Direction vector is (begin{pmatrix} 0.5 0.5 end{pmatrix}). Slope is 0.5/0.5 = 1. Perfect, matches the slope. So, Option E is correct.Wait a second, earlier I thought Option A was correct, but when I checked the initial problem, the user said that in their solution, A was incorrect. Did I make a mistake?Let me double-check Option A. The point is (-1, -2), which is on the line. The direction vector is (2, 2), which has a slope of 1. So, why would it be incorrect? Maybe I need to plug in the parameterization into the line equation to verify.For Option A:x = -1 + 2ty = -2 + 2tPlugging into y = x - 1:-2 + 2t = (-1 + 2t) - 1-2 + 2t = -2 + 2tThat's an identity, so it's correct. Hmm, so why did the initial solution say A was incorrect? Maybe I'm missing something.Wait, perhaps the direction vector needs to be a specific multiple? Or maybe the point is not on the line? No, (-1, -2) is on the line. Direction vector slope is 1. So, I think A is correct. Maybe the initial solution had a mistake.But in the initial solution, A was marked as incorrect because substituting led to a contradiction, but when I substitute, it works. Maybe they made a mistake in their substitution.Let me try substituting again:For Option A:x = -1 + 2ty = -2 + 2tSo, y = x - 1 becomes:-2 + 2t = (-1 + 2t) - 1-2 + 2t = -2 + 2tYes, that's correct. So, Option A should be correct. Maybe the initial solution had an error.But according to my calculations, A, B, C, and E are correct, while D is incorrect. However, the initial solution said B, C, E are correct. So, there's a discrepancy.Wait, maybe I need to consider the direction vector's direction. For example, in Option A, the direction vector is (2, 2), which is in the same direction as (1,1). So, it's just scaled by 2. That should still be fine because direction vectors can be scaled.Similarly, in Option E, the direction vector is (0.5, 0.5), which is just scaled by 0.5. So, that's also fine.So, I think A should be correct. Maybe the initial solution had a mistake in their substitution.But to be thorough, let me check all options again.**Rechecking Option A:**x = -1 + 2ty = -2 + 2tPlug into y = x -1:-2 + 2t = (-1 + 2t) -1-2 + 2t = -2 + 2tYes, it's correct.**Rechecking Option B:**x = 1 + ty = 0 + tPlug into y = x -1:t = (1 + t) -1t = tCorrect.**Rechecking Option C:**x = 0 - ty = -1 - tPlug into y = x -1:-1 - t = (0 - t) -1-1 - t = -t -1Correct.**Rechecking Option D:**x = 3 + 3ty = 2 + 4tPlug into y = x -1:2 + 4t = (3 + 3t) -12 + 4t = 2 + 3tWhich simplifies to 4t = 3t, so t = 0. That only holds for t=0, not for all t. So, D is incorrect.**Rechecking Option E:**x = 2 + 0.5ty = 1 + 0.5tPlug into y = x -1:1 + 0.5t = (2 + 0.5t) -11 + 0.5t = 1 + 0.5tCorrect.So, according to my rechecking, A, B, C, and E are correct, while D is incorrect. However, the initial solution only marked B, C, and E as correct. I think there might have been a mistake in the initial solution regarding Option A.But to be safe, let me consider if there's any other reason why A might be incorrect. Maybe the direction vector is not appropriate? But (2,2) is a valid direction vector for slope 1. So, I think A is correct.Therefore, the valid parameterizations are A, B, C, and E.

answer：Okay, so I need to find the locus of points ( X ) from which tangents can be drawn to a given arc ( AB ) of a circle. Hmm, let me start by recalling some basic concepts about circles and tangents.First, I remember that a tangent to a circle is a straight line that touches the circle at exactly one point. This point is called the point of tangency. Also, an important property is that the radius of the circle at the point of tangency is perpendicular to the tangent line. So, if I have a circle with center ( O ) and a tangent at point ( A ), then ( OA ) is perpendicular to the tangent at ( A ).Now, the problem is about finding all such points ( X ) from which tangents can be drawn to the arc ( AB ). Let me visualize this. Suppose I have a circle with center ( O ), and an arc ( AB ) on its circumference. I need to find all points ( X ) such that from ( X ), I can draw two tangents to the circle, each touching the circle at some point on arc ( AB ).Wait, is it two tangents or just one? The problem says "tangents," plural, so I think it's two tangents. So, each point ( X ) should be such that there are two distinct tangents from ( X ) to the circle, and both points of tangency lie on arc ( AB ).But hold on, if I consider the entire circle, the locus of points from which tangents can be drawn to the circle is the exterior of the circle. However, in this case, we are restricted to tangents that touch only the arc ( AB ). So, the locus should be a subset of the exterior of the circle, but specifically related to the arc ( AB ).Let me think about how to construct this locus. If I fix the circle and the arc ( AB ), then for any point ( X ) outside the circle, there are two tangents from ( X ) to the circle. The points where these tangents touch the circle are called the points of contact. For ( X ) to be in our desired locus, both points of contact must lie on arc ( AB ).So, how can I characterize all such points ( X )? Maybe I can use the concept of the polar line. The polar of a point ( X ) with respect to a circle is the line such that if ( X ) lies outside the circle, the polar is the line joining the points of contact of the tangents from ( X ) to the circle.Therefore, if the polar of ( X ) intersects the circle at points on arc ( AB ), then ( X ) lies in our desired locus. So, the locus of ( X ) is the set of all points whose polars with respect to the circle intersect the circle at points on arc ( AB ).But I'm not sure if this is the most straightforward way to describe the locus. Maybe there's a more geometric way to visualize it.Another approach is to consider the angles involved. For a point ( X ) outside the circle, the angle between the two tangents from ( X ) to the circle is related to the position of ( X ). Specifically, the angle between the tangents is equal to the angle subtended by the chord of contact at the center of the circle.Wait, the chord of contact is the line joining the points of contact of the tangents from ( X ). So, if the chord of contact intersects the circle at points on arc ( AB ), then the angle subtended by this chord at the center ( O ) is equal to the central angle corresponding to arc ( AB ).Let me denote the central angle for arc ( AB ) as ( theta ). Then, the angle between the tangents from ( X ) should be equal to ( theta ). But actually, the angle between the tangents is equal to the angle subtended by the chord of contact, which is related to the central angle.I think the angle between the tangents is supplementary to the central angle. So, if the central angle is ( theta ), then the angle between the tangents is ( 180^circ - theta ).Therefore, for all points ( X ) such that the angle between the tangents from ( X ) to the circle is ( 180^circ - theta ), where ( theta ) is the measure of arc ( AB ), those points ( X ) lie on our desired locus.But how do I describe this locus? It seems like it's a circular arc itself. Maybe the locus is another circle or an arc of a circle.Wait, I recall that the locus of points from which the angle between the tangents to a circle is constant is another circle concentric with the original circle. But in this case, the angle isn't constant for all points, it's specifically related to the arc ( AB ).Hmm, maybe I need to construct the locus by considering the two tangents from ( X ) to arc ( AB ). Let me try to think about the envelope of these tangents.Alternatively, perhaps I can use inversion geometry. Inversion might transform the problem into something simpler, but I'm not sure if that's necessary here.Let me go back to the basics. For any point ( X ) outside the circle, the set of points from which tangents can be drawn to the circle is the exterior of the circle. But we need to restrict this to points where both points of tangency lie on arc ( AB ).So, if I fix arc ( AB ), then the locus of ( X ) is the set of points such that the chord of contact from ( X ) intersects the circle at points on arc ( AB ).This chord of contact is the polar of ( X ) with respect to the circle. So, the condition is that the polar of ( X ) intersects the circle at points on arc ( AB ).Therefore, the locus of ( X ) is the set of points whose polars intersect the circle at points on arc ( AB ). This is equivalent to saying that the polar of ( X ) is a chord of the circle lying on arc ( AB ).But how do I describe this set of points ( X )? It seems like it's another circle or perhaps a region bounded by certain lines.Wait, maybe I can consider the two tangents from ( X ) to the endpoints ( A ) and ( B ). If I fix points ( A ) and ( B ), then the locus of points ( X ) from which tangents can be drawn to both ( A ) and ( B ) is the set of points lying on the external homothety axis of the two tangent circles at ( A ) and ( B ).But I'm not sure if that's the right approach. Maybe I need to think about the external and internal bisectors.Alternatively, perhaps the locus is the region outside the circle bounded by the two tangents at ( A ) and ( B ). So, if I draw the tangents at ( A ) and ( B ), the region between these two tangents outside the circle would be the locus of points ( X ) from which tangents can be drawn to arc ( AB ).Wait, that makes sense. Because if I draw the tangents at ( A ) and ( B ), any point outside the circle between these two tangents would have tangents that touch the circle somewhere on arc ( AB ). Conversely, points outside this region would have tangents that touch the circle on the opposite arc.So, the locus should be the region bounded by the two tangents at ( A ) and ( B ) and the circle itself. But actually, it's the exterior region bounded by these two tangents.But I need to be precise. Let me try to formalize this.Consider the circle with center ( O ) and radius ( r ). Let arc ( AB ) be a given arc on the circumference. The tangents at ( A ) and ( B ) are lines perpendicular to ( OA ) and ( OB ), respectively.The region from which tangents can be drawn to arc ( AB ) is the set of points ( X ) such that the angle between the tangents from ( X ) to the circle is equal to the angle subtended by arc ( AB ) at the center.Wait, earlier I thought the angle between the tangents is supplementary to the central angle. Let me verify that.If the central angle is ( theta ), then the angle between the tangents is ( 180^circ - theta ). Yes, that seems correct because the angle between the tangents is equal to the angle between the lines from ( X ) to ( A ) and ( B ), which is supplementary to the central angle.Therefore, the locus of points ( X ) is the set of points from which the angle between the tangents to the circle is ( 180^circ - theta ), where ( theta ) is the measure of arc ( AB ).But how do I describe this set geometrically? It seems like it's another circle. Specifically, the locus is a circle called the "circle of antisimilitude" or something similar.Wait, no, perhaps it's the circle with diameter ( AB ). But that doesn't sound right because the diameter would be a straight line, not a circle.Alternatively, maybe it's the circle passing through ( A ) and ( B ) and tangent to the original circle at some point. Hmm, not sure.Wait, another idea. If I fix points ( A ) and ( B ), then the locus of points ( X ) such that the angle ( AXB ) is constant is a circular arc. But in this case, the angle between the tangents is constant, not the angle at ( X ).Hmm, perhaps I need to use the concept of the director circle. The director circle of a circle is the locus of points from which the angle between the tangents to the circle is ( 90^circ ). So, in general, the locus of points from which the angle between the tangents is constant is a circle.Therefore, if the angle between the tangents is ( 180^circ - theta ), then the locus is a circle whose radius can be calculated based on the original circle's radius and the angle ( theta ).Let me recall the formula for the radius of the director circle. For a circle of radius ( r ), the director circle (where the angle between tangents is ( 90^circ )) has a radius of ( sqrt{2}r ).In general, for an angle ( phi ) between the tangents, the radius ( R ) of the locus circle is given by ( R = frac{r}{sin(phi/2)} ).So, in our case, the angle between the tangents is ( 180^circ - theta ). Therefore, ( phi = 180^circ - theta ), and the radius ( R ) is ( frac{r}{sin((180^circ - theta)/2)} = frac{r}{sin(90^circ - theta/2)} = frac{r}{cos(theta/2)} ).Therefore, the locus is a circle with radius ( frac{r}{cos(theta/2)} ) centered at the same center ( O ) as the original circle.Wait, but does this circle lie entirely outside the original circle? Let me check.Since ( cos(theta/2) ) is less than or equal to 1, ( frac{r}{cos(theta/2)} ) is greater than or equal to ( r ). So, yes, this circle is concentric with the original circle and has a larger radius, meaning it lies outside the original circle.But wait, this seems to suggest that the locus is a full circle, but in reality, we are only considering points from which tangents can be drawn to arc ( AB ). So, maybe only a portion of this circle is relevant.Alternatively, perhaps the locus is not a full circle but a specific region bounded by certain lines.Wait, going back to the initial idea, if I draw the tangents at ( A ) and ( B ), the region between these two tangents outside the circle is where the points ( X ) lie such that the tangents from ( X ) touch the circle on arc ( AB ).Therefore, the locus is the region bounded by the two tangents at ( A ) and ( B ) and the circle itself. But actually, it's the exterior region bounded by these two tangents.But I need to make sure. Let me consider a specific example. Suppose arc ( AB ) is a semicircle, so ( theta = 180^circ ). Then, the angle between the tangents would be ( 180^circ - 180^circ = 0^circ ), which doesn't make sense. Wait, that can't be right.Wait, no, if arc ( AB ) is a semicircle, then the tangents at ( A ) and ( B ) are parallel because the radius at ( A ) is perpendicular to the tangent, and similarly for ( B ). Since ( OA ) and ( OB ) are in opposite directions for a semicircle, their tangents are parallel.Therefore, the region between the two parallel tangents would be the locus of points ( X ) from which tangents can be drawn to the semicircle arc ( AB ). But in this case, the locus would be the region between the two parallel lines, which is an infinite strip.But according to the earlier formula, if ( theta = 180^circ ), then ( R = frac{r}{cos(90^circ)} ), but ( cos(90^circ) = 0 ), which would make ( R ) undefined. So, that suggests that the formula doesn't apply when ( theta = 180^circ ), which makes sense because the director circle concept breaks down when the angle is 180 degrees.Therefore, perhaps the locus is not always a circle. It depends on the measure of arc ( AB ). If arc ( AB ) is less than a semicircle, then the locus is a circle. If it's a semicircle, the locus is a region between two parallel lines.But in the general case, when arc ( AB ) is less than a semicircle, the locus is a circle with radius ( frac{r}{cos(theta/2)} ) centered at ( O ). However, this circle would encompass all points from which the angle between the tangents is ( 180^circ - theta ).But wait, in our problem, we are only considering points from which tangents can be drawn to arc ( AB ). So, even though the director circle gives us points where the angle between tangents is fixed, we need to ensure that both points of tangency lie on arc ( AB ).Therefore, perhaps the locus is not the entire director circle but only a portion of it. Specifically, the portion that corresponds to points ( X ) such that the chord of contact intersects the circle at points on arc ( AB ).Alternatively, maybe the locus is the region bounded by the two tangents at ( A ) and ( B ) and the circle. So, it's the area outside the circle but between the two tangents.Wait, let me think about this. If I have a point ( X ) outside the circle, and I draw tangents to the circle from ( X ), the points of contact will lie on the circle. If I want both points of contact to lie on arc ( AB ), then ( X ) must lie in a specific region.Imagine moving point ( X ) around. When ( X ) is close to the circle, the points of contact will be near ( A ) and ( B ). As ( X ) moves away, the points of contact spread out. However, if ( X ) moves too far in a particular direction, the points of contact might leave arc ( AB ).Therefore, the locus is bounded by the two tangents at ( A ) and ( B ). Any point ( X ) lying between these two tangents (outside the circle) will have tangents that touch the circle on arc ( AB ). Conversely, points outside this region will have tangents that touch the circle on the opposite arc.So, the locus is the region bounded by the two tangents at ( A ) and ( B ) and the circle. But since we are talking about points from which tangents can be drawn, it's the exterior region bounded by these two tangents.Therefore, the locus is the set of all points lying outside the circle and between the two tangents at ( A ) and ( B ).But to express this more formally, perhaps we can describe it as the region outside the circle bounded by the two tangents at ( A ) and ( B ).Alternatively, in terms of geometric construction, the locus can be described as follows:1. Draw the given circle with center ( O ) and the arc ( AB ).2. Draw the tangents at points ( A ) and ( B ).3. The locus of points ( X ) is the region outside the circle and between these two tangents.So, in summary, the locus is the region outside the circle bounded by the two tangents at ( A ) and ( B ).But wait, let me verify this with a specific example. Suppose arc ( AB ) is a minor arc, say 60 degrees. Then, the tangents at ( A ) and ( B ) will form a certain angle. Points ( X ) lying between these tangents will have tangents that touch the circle on arc ( AB ).If I move ( X ) beyond one of the tangents, say beyond the tangent at ( A ), then one of the tangents from ( X ) will touch the circle on the opposite side of ( A ), which is not on arc ( AB ). Therefore, ( X ) must lie between the two tangents to ensure both points of contact are on arc ( AB ).Therefore, my conclusion is that the locus of points ( X ) is the region outside the circle bounded by the two tangents at ( A ) and ( B ).But to express this in a more mathematical way, perhaps using equations.Let me set up a coordinate system. Let the circle have center at the origin ( O(0,0) ) and radius ( r ). Let points ( A ) and ( B ) be on the circumference, with coordinates ( A(r,0) ) and ( B(rcostheta, rsintheta) ), where ( theta ) is the central angle for arc ( AB ).The tangent at ( A ) is vertical, since the radius ( OA ) is horizontal, so the tangent is perpendicular to ( OA ). Therefore, the tangent at ( A ) is the line ( x = r ).Similarly, the tangent at ( B ) can be found. The slope of ( OB ) is ( tantheta ), so the slope of the tangent at ( B ) is ( -cottheta ). Therefore, the equation of the tangent at ( B ) is ( y - rsintheta = -cottheta (x - rcostheta) ).Simplifying this, we get:( y = -cottheta cdot x + rcostheta cdot cottheta + rsintheta )( y = -cottheta cdot x + rcostheta cdot frac{costheta}{sintheta} + rsintheta )( y = -cottheta cdot x + frac{rcos^2theta}{sintheta} + rsintheta )( y = -cottheta cdot x + frac{rcos^2theta + rsin^2theta}{sintheta} )( y = -cottheta cdot x + frac{r(cos^2theta + sin^2theta)}{sintheta} )( y = -cottheta cdot x + frac{r}{sintheta} )So, the tangent at ( B ) is ( y = -cottheta cdot x + frac{r}{sintheta} ).Now, the region bounded by these two tangents and outside the circle is the locus of points ( X ).But to describe this region mathematically, we can write inequalities. For the tangent at ( A ), ( x geq r ). For the tangent at ( B ), ( y geq -cottheta cdot x + frac{r}{sintheta} ).But actually, the region is between the two tangents, so it's the intersection of the regions defined by ( x geq r ) and ( y geq -cottheta cdot x + frac{r}{sintheta} ), but also outside the circle.Wait, no, because depending on the angle ( theta ), the tangents might intersect at some point, forming a sort of angle. So, the locus is actually the region outside the circle and between the two tangents.But to express this precisely, we need to consider the intersection of the two tangents. Let me find the point where the two tangents intersect.Set ( x = r ) in the equation of the tangent at ( B ):( y = -cottheta cdot r + frac{r}{sintheta} )( y = -rcottheta + frac{r}{sintheta} )( y = -rfrac{costheta}{sintheta} + frac{r}{sintheta} )( y = frac{-rcostheta + r}{sintheta} )( y = frac{r(1 - costheta)}{sintheta} )Using the identity ( 1 - costheta = 2sin^2(theta/2) ) and ( sintheta = 2sin(theta/2)cos(theta/2) ), we get:( y = frac{2rsin^2(theta/2)}{2sin(theta/2)cos(theta/2)} )( y = frac{rsin(theta/2)}{cos(theta/2)} )( y = rtan(theta/2) )So, the two tangents intersect at point ( P(r, rtan(theta/2)) ).Therefore, the locus of points ( X ) is the region outside the circle and between the two tangents, which form an angle at point ( P ).But how do we describe this region? It's the set of all points ( X ) such that ( X ) lies outside the circle and on the same side of both tangents as the arc ( AB ).In other words, the locus is the region bounded by the two tangents at ( A ) and ( B ) and the circle, excluding the interior of the circle.But to express this in terms of equations, we can say that for any point ( X(x,y) ), it must satisfy:1. ( (x - 0)^2 + (y - 0)^2 geq r^2 ) (outside or on the circle)2. ( x geq r ) (to the right of the tangent at ( A ))3. ( y geq -cottheta cdot x + frac{r}{sintheta} ) (above the tangent at ( B ))But actually, depending on the position of arc ( AB ), the inequalities might change. For example, if arc ( AB ) is in a different position, the tangents might be oriented differently.Alternatively, a more general way to describe the locus is that it is the region outside the circle and between the two tangents at ( A ) and ( B ).Therefore, the locus is the set of all points ( X ) such that ( X ) lies outside the circle and between the two tangents at ( A ) and ( B ).But to make this more precise, perhaps we can describe it as the region bounded by the two tangents and the circle, excluding the interior of the circle.In conclusion, after considering various approaches and verifying with specific examples, I believe the locus of points ( X ) from which tangents can be drawn to the given arc ( AB ) of a circle is the region outside the circle bounded by the two tangents at ( A ) and ( B ).

answer：Okay, so I have this problem here about a parabola, and I need to figure out the distance from a point M on the parabola to the y-axis. The parabola is given by the equation ( y^2 = 4x ), and the distance from M to the focus is 10. Hmm, let's see how I can approach this.First, I remember that for a parabola in the form ( y^2 = 4ax ), the focus is at ( (a, 0) ). In this case, comparing ( y^2 = 4x ) with the standard form, I can see that ( 4a = 4 ), so ( a = 1 ). That means the focus of this parabola is at ( (1, 0) ). Got that down.Now, the problem says that the distance from point M to the focus is 10. So, if I denote point M as ( (x, y) ), then the distance between ( (x, y) ) and the focus ( (1, 0) ) is 10. Using the distance formula, that would be:[sqrt{(x - 1)^2 + (y - 0)^2} = 10]Simplifying that, I get:[sqrt{(x - 1)^2 + y^2} = 10]If I square both sides to eliminate the square root, it becomes:[(x - 1)^2 + y^2 = 100]Alright, so that's one equation involving x and y. But I also know that point M lies on the parabola ( y^2 = 4x ). Maybe I can substitute ( y^2 ) from the parabola equation into the distance equation to solve for x.Substituting ( y^2 = 4x ) into the distance equation:[(x - 1)^2 + 4x = 100]Let me expand ( (x - 1)^2 ):[x^2 - 2x + 1 + 4x = 100]Combine like terms:[x^2 + 2x + 1 = 100]Wait, that simplifies to:[x^2 + 2x + 1 = 100]Hmm, that looks like a quadratic equation. Let me write it as:[x^2 + 2x + 1 - 100 = 0][x^2 + 2x - 99 = 0]Okay, so I have a quadratic equation ( x^2 + 2x - 99 = 0 ). I can solve this using the quadratic formula. The quadratic formula is ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = 2 ), and ( c = -99 ).Plugging in the values:[x = frac{-2 pm sqrt{(2)^2 - 4(1)(-99)}}{2(1)}][x = frac{-2 pm sqrt{4 + 396}}{2}][x = frac{-2 pm sqrt{400}}{2}][x = frac{-2 pm 20}{2}]So, that gives two solutions:1. ( x = frac{-2 + 20}{2} = frac{18}{2} = 9 )2. ( x = frac{-2 - 20}{2} = frac{-22}{2} = -11 )Wait a second, x can't be negative in this parabola, right? Because the parabola ( y^2 = 4x ) opens to the right, so x should be greater than or equal to 0. So, x = -11 doesn't make sense here. Therefore, the only valid solution is x = 9.So, point M has an x-coordinate of 9. The question asks for the distance from M to the y-axis. The distance from a point to the y-axis is simply the absolute value of its x-coordinate. Since x is 9, the distance is |9| = 9.Let me just double-check my steps to make sure I didn't make any mistakes.1. Identified the focus correctly at (1, 0).2. Applied the distance formula correctly, leading to the equation ( (x - 1)^2 + y^2 = 100 ).3. Substituted ( y^2 = 4x ) into the equation, resulting in ( x^2 + 2x - 99 = 0 ).4. Solved the quadratic equation correctly, obtaining x = 9 and x = -11.5. Discarded x = -11 as it's not valid for the parabola.6. Concluded that the distance to the y-axis is 9.Everything seems to check out. I think I'm confident with this answer.**Final Answer**boxed{D}

answer：Okay, so I have this problem about angles α and β in the Cartesian coordinate system. Their terminal sides intersect the unit circle at specific points, and I need to find the value of sinα cosβ. Hmm, let me try to figure this out step by step.First, I remember that in the unit circle, any point (x, y) on the circle corresponds to the cosine and sine of the angle formed with the positive x-axis. So, for angle α, the point is (12/13, 5/13). That means cosα should be 12/13 and sinα should be 5/13, right? Because x is cosine and y is sine.Wait, but hold on. The point given is (12/13, 5/13). Since both x and y are positive, that means angle α is in the first quadrant. So, both sine and cosine should be positive there. That makes sense. So, sinα is 5/13 and cosα is 12/13.Now, moving on to angle β. The terminal side intersects the unit circle at (-3/5, 4/5). Hmm, okay, so the x-coordinate is negative and the y-coordinate is positive. That should place angle β in the second quadrant. In the second quadrant, sine is positive and cosine is negative. So, sinβ would be 4/5 and cosβ would be -3/5.Wait, but the question is asking for sinα cosβ. So, I need to find sinα and cosβ and multiply them together. From above, sinα is 5/13 and cosβ is -3/5. So, multiplying those together: (5/13) * (-3/5). Let me compute that.First, 5 times -3 is -15, and 13 times 5 is 65. So, that gives me -15/65. Hmm, can I simplify that fraction? Let's see, both 15 and 65 are divisible by 5. So, dividing numerator and denominator by 5, I get -3/13.Wait, so is the answer -3/13? Let me double-check my steps to make sure I didn't make a mistake.Starting with point (12/13, 5/13) for α: yes, in the first quadrant, so sinα is 5/13 and cosα is 12/13. That seems right.For β, the point is (-3/5, 4/5). So, in the second quadrant, sinβ is positive 4/5 and cosβ is negative 3/5. So, cosβ is -3/5. That seems correct.Multiplying sinα and cosβ: (5/13) * (-3/5). The 5s cancel out, leaving -3/13. Yeah, that makes sense.Wait, but in the initial calculation, I got -15/65, which simplifies to -3/13. So, both ways, I get the same answer. So, I think that's correct.But just to be thorough, let me recall the definitions again. On the unit circle, for any angle θ, the coordinates are (cosθ, sinθ). So, for α, (12/13, 5/13) means cosα = 12/13 and sinα = 5/13. For β, (-3/5, 4/5) means cosβ = -3/5 and sinβ = 4/5.So, sinα is 5/13 and cosβ is -3/5. Multiplying them: (5/13) * (-3/5) = (-15)/65 = -3/13. Yep, that's consistent.I think I've got it right. So, the value of sinα cosβ is -3/13.**Final Answer**The value of sinalphacosbeta is boxed{-dfrac{3}{13}}.

answer：Okay, so I have this problem here: I need to find the constant term in the expansion of ((x^2 + frac{1}{x^2} - 2)^3). The options are A: -8, B: -12, C: -20, D: 20. Hmm, let me think about how to approach this.First, I remember that when dealing with expansions, especially binomial expansions, it's helpful to simplify the expression if possible. The given expression is ((x^2 + frac{1}{x^2} - 2)^3). I notice that (x^2 + frac{1}{x^2}) looks similar to a square of something. Let me check:If I consider ((x - frac{1}{x})^2), expanding that gives (x^2 - 2 + frac{1}{x^2}). Wait, that's almost the same as the expression inside the parentheses here, except the middle term is -2 instead of +2. So actually, (x^2 + frac{1}{x^2} - 2) is equal to ((x - frac{1}{x})^2). Let me verify:[(x - frac{1}{x})^2 = x^2 - 2 cdot x cdot frac{1}{x} + frac{1}{x^2} = x^2 - 2 + frac{1}{x^2}]Yes, that's exactly the same as (x^2 + frac{1}{x^2} - 2). So, the original expression can be rewritten as:[(x^2 + frac{1}{x^2} - 2)^3 = left( (x - frac{1}{x})^2 right)^3 = (x - frac{1}{x})^6]Alright, so now the problem simplifies to finding the constant term in the expansion of ((x - frac{1}{x})^6). That seems more manageable.I recall that the binomial expansion of ((a + b)^n) is given by:[sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k]In this case, (a = x), (b = -frac{1}{x}), and (n = 6). So, the general term in the expansion is:[binom{6}{k} x^{6 - k} left(-frac{1}{x}right)^k]Let me simplify that term:First, (left(-frac{1}{x}right)^k = (-1)^k cdot frac{1}{x^k} = (-1)^k x^{-k}).So, the general term becomes:[binom{6}{k} x^{6 - k} cdot (-1)^k x^{-k} = binom{6}{k} (-1)^k x^{6 - 2k}]We need to find the constant term, which means the exponent of (x) should be 0. So, set the exponent equal to 0:[6 - 2k = 0]Solving for (k):[6 = 2k implies k = 3]So, the constant term corresponds to (k = 3). Let's compute that term:[binom{6}{3} (-1)^3 x^{6 - 2 cdot 3} = binom{6}{3} (-1)^3 x^{0}]Simplify the binomial coefficient and the power of -1:[binom{6}{3} = frac{6!}{3! cdot (6 - 3)!} = frac{720}{6 cdot 6} = frac{720}{36} = 20]And ((-1)^3 = -1). So, the constant term is:[20 cdot (-1) = -20]Therefore, the constant term in the expansion is -20, which corresponds to option C.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I recognized that (x^2 + frac{1}{x^2} - 2) is equal to ((x - frac{1}{x})^2). That seems correct because when I expand ((x - frac{1}{x})^2), I get (x^2 - 2 + frac{1}{x^2}), which is the same as the given expression.2. Then, I rewrote the original expression as ((x - frac{1}{x})^6). That makes sense because (( (x - frac{1}{x})^2 )^3 = (x - frac{1}{x})^{6}).3. Using the binomial theorem, I set up the general term as (binom{6}{k} (-1)^k x^{6 - 2k}). That looks right.4. I set the exponent (6 - 2k = 0) to find the constant term, which gave me (k = 3). That seems correct.5. Calculating the binomial coefficient (binom{6}{3}), I got 20. Let me verify that:[binom{6}{3} = frac{6!}{3!3!} = frac{720}{6 cdot 6} = frac{720}{36} = 20]Yes, that's correct.6. Then, I multiplied by ((-1)^3 = -1), so the term is -20. That seems right.I think all my steps are correct. So, the constant term is indeed -20, which is option C.**Final Answer**boxed{C}