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A line of fixed length 2 units moves so that its ends are on the positive $x$-axis and that part of the line $x+y=$ 0 which lies in the second quadrant. The locus o...

Question

A line of fixed length 2 units moves so that its ends are on the positive $x$-axis and that part of the line $x+y=$ 0 which lies in the second quadrant. The locus of the mid-point of the line has the equation(A) $(x+2 y)^{2}+y^{2}=1$(B) $(x-2 y)^{2}+y^{2}=1$(C) $(x+2 y)^{2}-y^{2}=1$(D) none of these

A line of fixed length 2 units moves so that its ends are on the positive $x$-axis and that part of the line $x+y=$ 0 which lies in the second quadrant. The locus of the mid-point of the line has the equation (A) $(x+2 y)^{2}+y^{2}=1$ (B) $(x-2 y)^{2}+y^{2}=1$ (C) $(x+2 y)^{2}-y^{2}=1$ (D) none of these



Answers

The solid lies between planes perpendicular to the $x$ -axis at $x=-\sqrt{2} / 2$ and $x=\sqrt{2} / 2 .$ The cross-sections perpendicular to the $x$ -axis are
a. circles whose diameters stretch from the $x$ -axis to the curve $y=2 / \sqrt[4]{1-x^{2}}.$
b. squares whose diagonals stretch from the $x$ -axis to the curve $y=2 / \sqrt[4]{1-x^{2}}.$

So for this problem, we have two points X one. Why one and x two? Why two? And then the these points lie on the line. Why is equal to M X plus beef? So you want to find the distance between these two points and were asked to do it using the arc length formula. So the ark, like l is given by the integral from X one two x two of the square root of one plus, uh, prime of X squared DX. So, in our case, um, f prime of X. That's gonna be, uh, why prime, which is gonna be the derivative of M X plus B, which is equal to We also noticed that since we're on a line we can represent m we we know what the slope is because we have two points. So em is actually gonna be the same as why to minus why one divided by ex too minus X one that's actually what are derivative is so looking at that we can substitute into our Klink formula. Um, so we had the integral from x one x two of the square of one, plus, um, why to minus why one divided by X two minus X one in that quantity squared. So we're gonna want to do some manipulations to combine the one and our derivative squared. So do that. We're going to say, OK, one. Well, that's equal to X to minus X one divided by itself. Uh, sorry that those were actually going to be squared. Um, because we're actually looking at, uh, are derivative squared that we're gonna add by two minus. Why one? It's weird. Divided by X two minus X one squared We have the squared of the whole thing DX now x two x one y two by one These air particular values they're not variables. Um, so this whole expression under a square root is constant so we can pull it outside of indigo. We're gonna do that. We're also going to combine the two fractions while we're at it. Um, we're gonna get x two minus x one. It's weird. Plus white too. Minus why one squared divided by X two minus x one squared square the entire thing. That's good times. The integral of X one claims x two of one d. X. So, um, the anti derivative of one is X. So we need to, um, evaluate X from X to thio x one. Uh, And when we do that, we just get X to minus x one. So this is going to be equal to, um, expression under the square root that's gonna be multiplied by X to minus X one. So then the last thing we want to do is we see that inter fraction the denominator is something squared s so we could actually pull that out So the numerator will have r squared of X two minus x one quantity squared, plus white to minus. Why one quantity squared the denominator. We're just left with X two minus x one, and this is multiplied with X to minors X one and then this is nice because we can cancel those terms out. We're left with the squared of X two minus x one squared plus y two minus. Why one squared? I didn't notice that this is indeed the distance formula that we're used to seeing. That's your

Lits centered at negative for negative, too. And I've got a major access horizontally that is six and a minor access of four. I'm gonna piece this all together and figure out what my equation is. So here's each hears. Kay. I know that the major access is to a so to a will be six meaning a will be three and to be is my minor access. So that is going to be four. Which means be is too. I know that since it's got a longer horizontal axis, it's gonna be this format X minus, each squared over a squared plus why minus k squared over B squared equals one. So I'll go ahead and plug in the values I have. I'm gonna have X minus and negative four. So plus four squared over a squared A is three. So that's nine plus. Why minus k and K is and negative. So it's gonna be add to squared over B Square B is too. So that'll be four equals one. Well, that is just gonna be C

Well, everyone, this is Ricky. And today we're working on Problem 24. We're being told that we have a plane with the point situated on one one, and that we have a point on a plane that no matter where you move it, it will always be equidistant from this point. So if we're to think about the difference shapes of forms that we are no off the only the only type of graft that we know of that is always equal distance was to its central point, would be a circle. For example, if we looked at your lips, there are portions of an ellipse that air closer for their way. If we look at a line all points or further away from this point. So circles are the only graphical form, we have distance. So I hope this fit

So this problem the distance formula is as a


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