How do you find the points of an ellipse where the tangent equals 1?

[delriofi]

Homework Statement

If there is an an ellipse x^2/9 + y^2/16 = 1, and the slope of the tangent is dx/dy = -16x/9y, how do you find what points at which the slope of the tangent is 1? I have no idea how to answer this and I've been trying for like an hour. Can anyone help me?

Homework Equations

The Attempt at a Solution

 

[Hurkyl]

how do you find what points at which the slope of the tangent is 1?

At the risk of stating the obvious... you just solve the equation.

Can you post the approaches you're taking? It's hard to make helpful responses if you don't tell us what you've been trying.

 

[hunt_mat]

Use y=4\sqrt{1-x^{2}/9} Differentiate this and set it equal to 1 and find x.

 

[delriofi]

I'm not sure what you mean... I haven't taken math in 4 years and now I have to learn calculus as a prerequisite so there's probably something really obvious that I should already know but don't.

I differentiate that x^2/9 + y^2/16 = 1 has the derivative -16x/9y, and so if -16x/9y = 1 then I'm supposed to find at what points on the ellipse this is the case. But I don't know how to do that. So that's all I did...

 

[jbunniii]

Well, the answer will be any point (x,y) that satisfies both equations:

x^2/9 + y^2/16 = 1
-16x/9y = 1

Solve the second equation for either x or y, and then substitute into the first equation.

 

[ckutlu]

You can not differentiate the equation directly like that since x and y are dependent on each other. What hunt_mat pointed out was, you should first put y to the one side of the equation then take the derivative of y(x) function for x. Your method for taking a derivative in a closed form is wrong. If you do what hunt_mat suggested, you'll get an equation in terms of x and after that only thing you need to do is to assign that equation to 1, then solve it by root finding process.

 

[delriofi]

I used implicit differentiation, so it still should have worked.

 

[jbunniii]

I used implicit differentiation, so it still should have worked.

Implicit differentiation is fine; I got the same equation as you did. The rest is easy! Tell us what you get if you solve

-16x/9y = 1

for x in terms of y. (Or y in terms of x. Your choice.)

 

[delriofi]

I get x = -9y/16 and y = -16x/9. So I should just do... x^2/9 + (-16x/9)^2/16 = 1? Or is that wrong?

 

[delriofi]

I keep getting that x^2/9 - x^2/9 = 1 but that don't make sense does it?

 

[jbunniii]

I keep getting that x^2/9 - x^2/9 = 1 but that don't make sense does it?

You must be making a mistake when substituting

x = -9y/16

into

x^2/9 + y^2/16 = 1

If x = -9y/16

then what is x^2/9?

 

[delriofi]

Is it -y^2/16?

 

[hunt_mat]

Following on from my suggestion, the derivative is:
<br /> \frac{dy}{dx}=\frac{4x}{9\sqrt{1-x^{2}/9}}=1<br />
Just re-arrange to find x, then insert this into the equation to find y.

 

[Hurkyl]

I get x = -9y/16 and y = -16x/9. So I should just do... x^2/9 + (-16x/9)^2/16 = 1? Or is that wrong?

That looks fine. What was your work from here? I don't see how you keep getting what you say below:

I keep getting that x^2/9 - x^2/9 = 1 but that don't make sense does it?

P.S. a gentle reminder for homework helpers: helping a person do a problem doesn't mean to make them abandon anything they were doing and do it the exactly the way you would do the problem. There are lots of way to solve a problem, and 9 times out of 10 it's better to help the person carry their own idea to the end.

 

[jbunniii]

Is it -y^2/16?

No, that's not right. Let's take a step back.

If x = -9y/16, then what is x^2?

 

[delriofi]

How do you solve (-16x/9)^2/9 ? That's some confusing algebra. I get -x^2/9 but that's wrong. What do you get?

 

[delriofi]

No, that's not right. Let's take a step back.

If x = -9y/16, then what is x^2?

Is it -9y^2/16 ?

 

[jbunniii]

How do you solve (-16x/9)^2/9 ? That's some confusing algebra. I get -x^2/9 but that's wrong. What do you get?

Do it in two steps.

First, you square (-16x/9).

Then, divide the result by 9.

Hint: the square can't be negative.

 

[delriofi]

In response to jbunniii, is it -9y^2/16 or no is it -9y^2/16^2 ?

 

[jbunniii]

Is it -9y^2/16 ?

No, this is where your mistake is. You aren't squaring properly.

If I have, say, (3*5) and I want to square it, then that's (3*5)^2 = (3*5) * (3*5) = (3*3) * (5*5) = 3^2 * 5^2.

Thus if you square the product of two things, the result is the product of the squares of those two things.

Same thing for quotients.

So you need to square each of the three components: -9, y, and 16.

 

[delriofi]

So it would be 81y^2/256 ?

 

[jbunniii]

So it would be 81y^2/256 ?

Right. So that's x^2. What is x^2/9?

 

[delriofi]

It would be... 81y^2/2304 I think?

 

[delriofi]

So...y = 0.110679 ?

 

[jbunniii]

It would be... 81y^2/2304 I think?

Right, or you could simplify it by factoring a 9 out of both 81 and 2304, leaving you with

x^2/9 = 9y^2/256.

Now substitute this into the ellipse equation

x^2/9 + y^2/16 = 1

and tell us what you get.

 

[jbunniii]

So...y = 0.110679 ?

No, that's not what I get. Please show us your calculation step by step so we can see what's wrong.

 

[delriofi]

Oh ok no sorry... 9y^2/256 + y^2/16 = 1 so 25y^2/256 = 1 and thus 25y^2 = 256 so... y = sqrt (256/25) which is 3.2 apparently... Is that better?

 

[hunt_mat]

That's what I get. You can get an exact number for that though, 256=16^2.

 

[jbunniii]

Oh ok no sorry... 9y^2/256 + y^2/16 = 1 so 25y^2/256 = 1 and thus 25y^2 = 256 so... y = sqrt (256/25) which is 3.2 apparently... Is that better?

Yes! Just one subtle point: 3.2 (or 16/5) is one number that squares to 256/25. There's also a second answer. What is it?

 

[hunt_mat]

My advice would be to draw a picture...

 

[delriofi]

Hahaha uhhh I guess the negative of 3.2, so -3.2 ? Good point. So because it's an ellipse then there are necessarily 2 points where dy/dx = 1 and so my answer should be like (4.5, -3.2) and (-4.5, 3.2) eh?

 

[hunt_mat]

Yes, that is right.

 

[delriofi]

Hurray! Thanks errbody. You guyz are so smart!