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Find the lines that are tangent and normal to the curve at the given point.2y2 = 100,The line tangent to the curve x2y2 = 100 at (5, 2) isy =The line normal to the ...

Question

Find the lines that are tangent and normal to the curve at the given point.2y2 = 100,The line tangent to the curve x2y2 = 100 at (5, 2) isy =The line normal to the curve x2y2 = 100 at (5,- 2) is y =

Find the lines that are tangent and normal to the curve at the given point. 2y2 = 100, The line tangent to the curve x2y2 = 100 at (5, 2) isy = The line normal to the curve x2y2 = 100 at (5,- 2) is y =



Answers

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. $$x^{2}=-10 y,(2 \sqrt{5},-2)$$

We want to find the lines that are tangent and normal to the curve at the given point. This is going to be using implicit differentiation. When we take the derivative, we get two x minus one over to root three X. Y. But then we also want to do the chain rule. So it's going to be times three um times Y plus X. Y. Prime. And then lastly we'll have plus for why? Why pray? So before we do anything else we know this is going to equal zero. And then we can just plug in our X and Y values and solve for Y prime. I would advise for you to plug in the X and Y values first because then it will get it more so in terms of simple constant values.

With this problem we have X squared plus y squared is equal to 25. The first thing you want to do is find the derivative. So using implicit differentiation and the power rule and you get to y dy dx there And then the drift of of 25 would be zero. So what you want to do is solve for dy dx And it would just simply be subtract two X over and divide by two Y. To cancel those 2s. Now, what we really want is the derivative dy dx at the ordered pair that gave us Which was 3 -4. So as I plug in three in for X I get negative three over negative four. So what we just found is the slope of the tangent. So alright that we tangents of is equal to positive three force. And the reason why I have to say that that's important is because when I asked for the line that is tangent And you use point so for maybe 3/4 X minus the X coordinate from the problem. And then close your parentheses minus the Y coordinate plus the white corn I guess I should say. So that's why I say is negative for so that's the tangent line and I would not simplify it because it's not helpful. Um In the normal line it's just the slope is perpendicular. So it's the negative reciprocal And it's still through the same point. So X -3 and the same Y coordinate. I will point out in a problem like this. The normal line should pass through the origin. So if you simplify that correctly, it will simply be negative 4/3 X if you wanted to.

So we are looking at why equals to sign of pi x minus Y. And the first thing you want to do is take the derivative of this. So the left side of the drift of wise do Iranians on the right side of the chain role because the derivative of sine is co sign and what you do is you leave what's inside the functional alone and then you multiply by the derivative of the inside which would be pi minus dy dx. So as far as solving for dy dx at the ordered pair Of What Is It? 10. What I would do is not even try to sulphur Dy Dx. I would just leave things alone And if I plug in one and 4 this piece will be to co sign of pie and plug in zero for why? Whatever my answer is there I need to multiply by um hi minus dy dx. I think it's easier to solve that way because if you think about what cosign a pie is Pies over here in the Unit Circle Co signs the excrement. So it's -1. So therefore dy dx is equal to two times negative ones have been negative two times the quantity of high minus dy dx. So if I distribute in there I would have negative too high plus to delete dx and then I can combine my terms by subtracting over. So I have a negative one Dy dx. So one dy dx minus two dy dx and it was negative two pi. And so I can figure out the slope after I divide both sides by name, one would be positive to pie. So the tangent line would be a. Y equals the slope we just found and then uh x minus the X according the problem. And you could write plus the Y coordinate but adding zero will not change the value of that. So this is a good answer. You might have A teacher prefer slope intercept form where you would just distribute in there. So in normal line, the only thing that's different for a normal line is it's perpendicular to the tangent, so the slope is the negative or super. So just flip too high And it's still going through the same point. So X -1. And again you can distribute if you want.

And this problem, we're given the implicit curve X squared minus squared of three x y plus to widespread equals five. And we're first as a verified that the point squared of three comma two lives on the curve. So to do that, we should plug in X equals squared of three. Why equals two to see if we get a valid solution or not? So if we get something like five equals five, and that means that the point lies on the curve. But if we give something incorrect like five equals four, then that means that we're not on the curve. Let's go ahead and plug things in squared of three squared minus square to three times the square root of three times too, plus two times two squared. So we're asking if this is equal to five. This is a question mark, so it's three minus square two, three times greater free is three times two is six plus two times two squared is eight. You're asking if this is equal to five on. In fact, three minus six plus eight is equal five. So the point does lie on the curve. Now we want to start talking about slopes. So we know that the slope of the tangent line is always going to be equal. The D Y d X. So the real question are there. I mean, the next question is, what about the slope of the normal line? And the property of the normal line that we care about is that is perpendicular to the tangent line. So to find the slope of a perpendicular line, you take the negative, reciprocal, negative one divided by the slope of the tangent line. So later on, once we get the slope of the tangent line will be able to figure this out. All right, so let's jump into it. We need to find the Y d. X. So we need to do some kind of implicit differentiation. So let's do that. Over here I start taking derivatives ddx off X squared Maya spirit of three acts. Why? Plus two y squared equals five. So the derivative of X squared is too X Now, the next term, the squared of three is just a constant. So that that can kind of get pulled out. No. And now we need a different she X plus y Is this is the product rule, so it's through the first time, the second plus the first times the derivative of the second. Finally, the Druid of two y squared is for Why D. Y. D. X. So this is an example. Our specific calculation. We worked on a lot of examples before, and finally they drew the five zero Let's expand things and and see kind of how the dust settles so to x minus Route three. Why minus route three times d y d x Times x time T Y d X plus four. Why d Y d X equals zero Let's keep all the D Y d X is on one side and let's factor out a d Y d x So on the left hand side we're going to have these two terms. So factor out of D Y D X are left with four Y minus route three times X And then on the right hand side, we're gonna have to subtract these two turns over. So root three. Why minus two X Now, this is enough to get us do idea, because we can just divide by this expression Over here we're left with fruit. Three times why minus two times X over four times why minus squared of three times x So here is D Y d X. So now we can use it to find the slope of the tangent line. So we just need to plug in r Point X equals square two three. Why equals two into our expression for D y d X? So do I. D X. Yeah, squared of three comma too. So we get through with grade of three times, too. So square three times. Why? Minus two times X minus two times Route three over four times to minus route three times, Route three. Well, we can see here that we're gonna get a zero on the numerator because we have to root three minus two, Route three and the denominator is eight. Minus three is five zero divided by five zero. So because the slope zero this tells us that we have a horizontal line because the slope is zero. So this is a horizontal line at why equals two. So the equations just y equals two. So it's this nice for is on the line. This was part es for the tangent line for part B to calculate the normal line. Well, the Taejin line was horizontal and the normal line is going to be perpendicular. So is perpendicular and perpendicular to horizontal line is a vertical line. So is the verdict groups. The vertical line at the point or to me at X equals square two three, So the vertical line it's just given by X equals the square root of three.


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