In Exercises 83–88, find equations for the lines that are tang...

Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.

xy + 2x - 5y = 2, (3, 2)

Accepted Answer

Welcome back, everyone. In this problem, we want to find the equations for both the tangent and normal lines to the curve 2 X Y minus 3X + 4 Y equals 7 at the 0.-31. A says it's the tangent line is Y equals 1/2 x minus 1/2, and the normal line is Y equals -2 X + 7. B says they are Y equals 1/2 minus 1/2 x and 2 X minus 7. C says they are negative x minus 1/2 and 2 X + 7, and D says they are 2 X + 7 and 1/2 minus 1/2 X. Now how can we find the equations for both the tangent and the normal lines? What do we know about the equation of a line? Well, recall that we can use the point slope form of a line to find its equation, which is Y minus Y1 equals M multiplied by X minus X1. In this problem, regardless of whichever line we're considering, we know we're looking at it at the 0.-31. So that means Y1 equals 1, and we can let X1 be equal to 3. No, that just means we need to figure out M. We need to figure out the tangent for the, for the, sorry, we need to figure out the slope for the tangent and for the normal. Let's start with the slope of the tangent. What do we know? We'll recall that the slope of the tangent line is going to be the derivative of or curve at that point. So let's differentiate our equation, OK? And then let's solve for the slope and plug it into the slope point form to get the equation of the tangent line, OK? So differentiating. To solve for him. OK. At -31, that means we're not gonna find the derivative of XY minus 3X plus 4 Y equals 7. And to find that derivative we can differentiate each of these with respect to x. So now when we do that, OK. We're going to end up with 2 Y plus 2XDYDX. This would be the derivative of 2 XY minus 3 plus 4 DYDX, the derivative of 4Y, OK. And the derivative of 7 equals 0. Now, we can solve for our derivative DUIDX, OK? So now, let's do a bit of rearranging here. Uh, let's combine our like terms, 2 X plus 4, sorry, 2 X DUIDX. Plus 4 DIDX. OK. And then let's move our other two terms to the other side of our equation. That's 3 minus 2y. And now if we factor 2 X + 4 and divide by that, then we should get our derivative to be equal to 3 minus 2y divided by 2 X + 4. Know that we have our derivative, we can find a slope at -31 by substituting X for -3 and Y for 1, OK? So when we do that, no. Or slope M, yeah. It is gonna be equal to 3 minus 2 multiplied by 1 divided by 2 multiplied by -3 + 4, which equals -2. Therefore, the slope of the tangent line is negative by 2. Now, since for our tangent line, OK, let me just add that here. OK. And for the tangent. Where we found that it's slope M equals -2. OK. We know that X equals -3, X1 is -3, sorry, and Y1 is 1, OK. Then by the slope point equation, by the slope point form of an equation, then we could write this as Y minus 1. Is going to be equal to -5 multiplied by X + 3. And if we simplify that, then we should get Y to be equal to -2 X minus 1/2. So this is the equation of the tangent line. Now what about the equation for the normal to the tangent? So what do we know about the normal to the tangent? Well, we know that the normal line to the tangent line or the normal is going to form a 90 degree angle to the tangent line. And when two lines are at 90 degree angles, in other words, when they are perpendicular, then the slope of one is the negative reciprocal of the other. So that means the slope of the normal, let's call it MN, is going to be the negative reciprocal of negative a half, which is going to be equal to 2. Since they use the same, since they're both going through the same points where X1 is -3 and Y1 equals 1. Then that means. Our equation is going to be Y minus 1 equals 2 multiplied by X + 3. And now if we simplify that, then we get the equation to be Y equals 2 x plus 7. Therefore, the equation of our tangent is Y equals Y equals -2 x minus 1/2, and the equation of the normal line is Y equals 2 X + 7. If we look back at our answer choices, that tells us C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.

xy + 2x - 5y = 2, (3, 2)

Key Concepts

Implicit Differentiation

Tangent Line

Normal Line

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