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(a) Suppose $ f $ is a one-to-one differentiable …

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Problem 76 Hard Difficulty

Find equations of both the tangent lines to the ellipse $ x^2 + 4y^2 = 36 $ that pass through the point (12, 3).


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03:50

Frank Lin

03:33

Doruk Isik

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Related Topics

Derivatives

Differentiation

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Lectures

Video Thumbnail

04:40

Derivatives - Intro

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

Video Thumbnail

44:57

Differentiation Rules - Overview

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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Video Transcript

So we're given um an ellipse and we're gonna find the two tangent lines that pass through the 20.12 3. Which is not on the ellipse, as you can see by the picture there. So what we're looking for is two lines. This one looks like it's probably gonna be a horizontal. And then the 2nd 1 that looks like it'll pass somewhere over here. And what we're gonna do is find the equations of these lines. And we want to do first is find the slopes and then we have to find out what these points are on these, on the ellipse that are actually part of the ellipse. Um to find the equation of the line. So in order to find the slope it's gonna be the derivative of this function. Um why prime? Right? So we need to find this, why prime. We're gonna differentiate, differentiate this implicitly. Um So basically we'll we'll treat excess like we normally do. And then anywhere we have a Y. It's gonna get a white crime attached to it. Alright, so derivative of X square is just two X. Derivative of four. Y squared is eight Y. But then because it's with respect to why we get a Y. Prime And then the derivative of 36 0. Now what I wanna do is I want to solve for Y. Prime because I want to find the slope. Right? So I want to isolate Y. Prime here. The way we'll do that is we subtract two X from both sides and we get eight Y Y. Prime equals negative two X. Viable size by Y. And we get why prime equal to negative X over four Y. After we simplify. And I know that that's also the slope of the line. Now I also know that I can find the slope using just the regular slope formula Y two minus Y one over X two minus X one. And I know my point here is 12 3. And I know that some equation at some point on the lips would be X one, Y one or X two, Y two. And so we'll just call this X one comma Y one. So then I know that this is gonna be equal to Y 2-. And I'll just call Let's just call it some random y -3 Because that's my one Over X -12. And I know that that's also equal to my expression that I got when I saw for Y prime. So that's equal to negative X over for why. And now what we can do is cross multiply here because I have two expressions that are fractions that are set equal. And we wind up with getting four y Times Why -3 equals negative X times x minus 12. So this is four Y squared minus three, Y equals negative X squared plus 12 X. And the trick to this is I know this expression here equals 36. So if I can find an X squared plus four Y squared in my equation down here, I can replace it with 36. And I can in fact do that if I add X squared to both sides here so we get four Y squared plus X squared minus three Y equals 12 X. Okay um This should be 12 y. No 12. Mhm. Was Yeah and and then I can now replace this piece here with 36 because that's my lips equation minus 12 Y equals 12 X. And now what I wanna do is I want to solve for X. So I can then plug it into my ellipse equation. So I'll divide both sides by 12. And this site simplifies to three -Y. X. Mhm. Okay. Absolutely. Mhm. Now if I go to our lips equation because I want to find a point on that ellipse four Y square plus X square Equals 36. And I replaced the X. with the expression on the left. We have now four Y squared plus three minus Y squared equals 36. We're four Y squared plus nine minus six. Y plus Y squared 36. Now this is quadratic. So what we wanna do is collect everything on the left. Set it equal to zero. So we get five Y squared minus six. Y. Uh plus -27 equals zero. And then we're going to use the quadratic formula. So your Y equals um opposite B plus a minus square root B squared minus four A. C. Q. A. Alright so that's gonna give us um six plus or minus square root of negative six squared minus four times five negative 27 Over two times 5. And this simplifies to six plus or minus square root 5 76/10. Thanks. which gives us to solutions like was three And -1.8. Okay. Oops. Okay so now we want to find the X. Is associated with that. So if we go back over here three minus Y equals X. So if I plug in Y is 3, -3 is equal to zero. So then I know the .03 is one of the points on my ellipse And then 3 -1.8 equals x. So that gives us 1.2 or 3 -2.8. Thank you. So that gives me 4.8 then. So then I know the .4.8 1.8 is also on the ellipse. And now to find the equation we can just use y minus y. One equals X -X. one. And I know from earlier m This equals um negative x over four y. And we'll just plug both of these points in. So if we use 03 We have Y -3 equals negative x 0/4. Y is times three X minus zero. This whole side becomes zero. So just get y minus three equals zero Which gives us y equals three. So that's the horizontal one that looked horizontal when we looked at the picture this is why it was three and then the other line um we're gonna use The .4.8. 1.8. So we have Y -1.8 equals negative 4.8 over 1.8 X minus. Is this negative? 1.8? This should be negative 1.8. So then this will actually change this two plus and that's a negative there here And 4.8 divided by 1.8 is 2.67. So let's make that a fraction 8/3. So this is y plus 1.8 equals eight thirds x minus 4.8. It looks like I forgot for here. It's back up here. Getting so there I forgot the four and the denominator for the um The soap here so it's just 4.8 negative over negative 1.8. Now the formula and it has a four by that by four. so this simplifies to 2/3 when we do that. Okay? Um So now let's simplify this. This is why plus 1.8 equals two thirds x two thirds times 4.8. There's 16/5 And then we'll add 1.8 to both or subtract 1.8 from both sides. Okay And we get y equals turn that into a fraction 7/5 so every 2/3 x minus seven fifths. Mhm. Right. Check that. I Put the wrong sign in my calculator -16 -1.8 is -5. Alright, so the two equations then after this long and winding problem here, Or y equals three. Which again was the horizontal, NY equals 2/3 x minus five, which if we can go back to the picture is the second line here And it looks like it goes through and I get a five pretty nicely there.

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Top Calculus 1 / AB Educators
Catherine Ross

Missouri State University

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University of Michigan - Ann Arbor

Joseph Lentino

Boston College

Calculus 1 / AB Courses

Lectures

Video Thumbnail

04:40

Derivatives - Intro

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

Video Thumbnail

44:57

Differentiation Rules - Overview

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

Join Course
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