Normal lines parallel to a line Find the norm...

Question

Normal lines parallel to a line Find the normal lines to the curve xy + 2x – y = 0 that are parallel to the line 2x + y = 0.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Identify all normal lines to the curve, Y equals X Y plus 3 X is equal to 0 that are parallel to the line, X minus 3 Y minus 1 is equal to 0. Awesome. So it appears our final answer we're ultimately trying to solve wars we're trying to determine all the normal lines that are normal to this curve for Y minus XY plus 3X equals 0. That is also parallel to the line. X minus 3 Y minus 1 is equal to 0. So now we know what we're ultimately trying to solve for. First off, we need to recall and note that a normal line to a curve at a given point is the line that is perpendicular to the tangent line at that point. So that means, therefore, we need to determine the slope of the line. X minus 3 Y minus 1 is equal to 0. So what we need to do is we need to solve for M, so we need to rearrange our expression so we can write -3 Y is equal to negative X plus 1. So rearranging this expression to isolate sulfur Y, we will find that Y is equal to 1/3 X minus 1/3. So that will mean that our slope M is equal to 1/3. Fantastic. So once again, we're just rearranging our initial equation for the line to rearrange it to isolate and solve for Y. and when we solve for Y, we're easily, since we have it in linear form now, we're easily able to identify what the slope is. So now our next step, we need to note that we need to recall a note that since the normal line is parallel to the line x minus 3 Y minus 1 is equal to 0. Its slope is also equal to 1/3. So as we should recall, once. Again, and even further, the slope of the normal line to the curve is the negative reciprocal of the slope of the tangent line. So that means that M perpendicular is equal to 1 divided by M10. So, So once again, the perpendicular slope is equal to 1 divided by the slope of the tangent line. So that means that M10, which once again, the the slope of the tangent line, is equal to -1 divided by M perpendicular, which is going to be equal to -1 divided by 1/3, which is going to be equal to -3. Fantastic. So the derivative of the function at a given point, as we should recall, represents the slope of the tangent line to the function at that point. So now, at this point, we need to differentiate both sides of the equation of the given curve with respect to X. So we need to take Dy by D X minus parentheses X multiplied by Dy by DX plus Y + 3 is equal to 0. So now, our next step is we need to take DY by DX minus X multiplied by DY by DX minus Y + 3 is equal to 0. Which means when we continue to solve, you will get parentheses 1 minus X multiplied by Dy by DX is equal to Y minus 3. So that will mean that DY by D X is equal to Y minus 3 divided by 1 minus X, which will mean that M10 is equal to Y minus 3 divided by 1 minus X. Fantastic. So now, our next step is we need to solve for Y in terms of X, so we need to take Y minus 3 divided by 1 minus X is equal to -3. So we need to rearrange this expression to isolate and solve for Y, so I'll skip a few steps, but we will find that Y is equal to 3 X. So all you need to do. I use algebra to rearrange our expression to isolate and solve for Y, and when we do that, we will find that Y is equal to 3X. And now we need to take 3 X. And we need to substitute this Y equals 3 X into our equation of the curve to help us find the X coordinate for the point of tangency. So we need to take Y minus XY plus 3 X is equal to 0, and then we can go ahead and write that 3X minus X multiplied by 3 X plus 3 X is equal to 0. So now what we need to do is we need to take this big old long expression, and we need to simplify to its most simple form and rearrange it to isolate and solve for X. And when we do that, we will find that X is simply equal to 2, which I skipped a few steps, but essentially all you're doing is you're just rearranging the expression to isolate and solve for X, and to give you a hint, when you start rearranging, you'll eventually find that it's going to be -3 X is equal to 0. Which will give us X is equal to 0, and you'll also find that we have X equals 2 because you'll find that X minus 2 is equal to 0. Awesome. So with that said, we now need to find the Y coordinates using the relationship Y equals 3 X. So for when X is equal to 0, let us note that Y is equal to 3 multiplied by 0, which will mean that Y is equal to 0. So it's going to be the coordinate point 0.0, which is the origin point. And for when X is equal to 2, that will mean that Y is equal to 3 multiplied by 2, which will mean that Y is equal to 6. Which will mean that we have a coordinate point of 2.6. So that will mean that the points of tangency are 0.0 and 2.6, both in parentheses. So now, our next step now is we need to find the equation of the normal lines using the point slope form, which, as we should recall, the point slope form states that Y minus Y1 is equal to M perpendicular multiplied by X minus X1. Awesome. So, that means, so, once again, that means that moving right along here, that our next steps, so for the condition, 0.0 in parentheses, and when M perpendicular perpendicular is equal to our slope for the perpendicular is going to be equal to 1/3. That will mean when we plug in our known variables that Y minus 0 is equal to 1/3 multiplied by X minus 0, so we rearrange this expression. To solve, like they get so when we rearrange this expression to Potentially get so we we're trying to move everything over so we get 0 on one side of the of the equal sign, and when we do that, we will find that X minus 3 Y is equal to 0. So what we're doing is we're taking our point slope form and we're rearranging our Y minus 0 equals 1/3 multiplied by X minus 0, and we're trying to move all of our variables to one side of the equal sign and get it to set it equal to 0. And when we do that, we will get that X minus 3 Y is equal to 0, and that is one of our final answers, that's one of our normal lines, but we need to consider our other point of tangency. So for our parentheses 26, and when our Slope for the perpendicular is also equal to 1/3, so that will mean that Y minus so once again it's going to be Y minus 6 is equal to 1/3 multiplied by X minus 2. So once again, when we rearrange our expression to have it set, so to get all of our variables onto one side of our equal sign and set it equal to 0, you will find that our final answer is going to be X minus 3Y + 16 is equal to 0, and that's it. That is our 2nd and final answer. Hooray, we did it. So ultimately, it's pretty simple at this stage. Once you get to the point slope form stage, all you need to do is just use a little bit of algebra and that and it's simple to solve. So that's it. So let's look at our problem itself and quickly recap all that we've learned together. So that means that our final two answers, once again, without a doubt, for once again, the normal lines to the curve that are parallel to the specific line. Are as follows. So, X minus 3 Y is equal to 0, and X minus 3Y + 16 is equal to 0. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Normal lines parallel to a line Find the normal lines to the curve xy + 2x – y = 0 that are parallel to the line 2x + y = 0.

Key Concepts

Normal Line

Implicit Differentiation

Parallel Lines

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