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

In summary, when given an ellipse with the equation x^2/9 + y^2/16 = 1 and the slope of the tangent being dx/dy = -16x/9y, the points at which the slope of the tangent is 1 are x = 3.2 and x = -3.2. This can be found by solving the equation -16x/9y = 1 and substituting the solutions into the ellipse equation.
  • #1
delriofi
17
0

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

 
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  • #2
delriofi said:
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.
 
  • #3
Use [itex]y=4\sqrt{1-x^{2}/9}[/itex] Differentiate this and set it equal to 1 and find x.
 
  • #4
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...
 
  • #5
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.
 
  • #6
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.
 
  • #7
I used implicit differentiation, so it still should have worked.
 
  • #8
delriofi said:
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.)
 
  • #9
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?
 
  • #10
I keep getting that x^2/9 - x^2/9 = 1 but that don't make sense does it?
 
  • #11
delriofi said:
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?
 
  • #12
Is it -y^2/16?
 
  • #13
Following on from my suggestion, the derivative is:
[tex]
\frac{dy}{dx}=\frac{4x}{9\sqrt{1-x^{2}/9}}=1
[/tex]
Just re-arrange to find x, then insert this into the equation to find y.
 
  • #14
delriofi said:
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:
delriofi said:
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.
 
  • #15
delriofi said:
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?
 
  • #16
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?
 
  • #17
jbunniii said:
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 ?
 
  • #18
delriofi said:
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.
 
  • #19
In response to jbunniii, is it -9y^2/16 or no is it -9y^2/16^2 ?
 
  • #20
delriofi said:
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.
 
  • #21
So it would be 81y^2/256 ?
 
  • #22
delriofi said:
So it would be 81y^2/256 ?

Right. So that's x^2. What is x^2/9?
 
  • #23
It would be... 81y^2/2304 I think?
 
  • #24
So...y = 0.110679 ?
 
  • #25
delriofi said:
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.
 
  • #26
delriofi said:
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.
 
  • #27
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?
 
  • #28
That's what I get. You can get an exact number for that though, 256=16^2.
 
  • #29
delriofi said:
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?
 
  • #30
My advice would be to draw a picture...
 
  • #31
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?
 
  • #32
Yes, that is right.
 
  • #33
Hurray! Thanks errbody. You guyz are so smart!
 

1. How do you define an ellipse?

An ellipse is a geometric shape that is formed by the intersection of a cone and a plane. It is characterized by having two focal points, where the sum of the distances from any point on the ellipse to the two focal points is constant.

2. What is the equation for an ellipse?

The standard equation for an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) represents the center of the ellipse and a and b are the lengths of the semi-major and semi-minor axes, respectively.

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

To find the points of an ellipse where the tangent equals 1, you can use the equation tan(θ) = b/a, where θ is the angle between the tangent line and the horizontal axis. Then, you can solve for the x and y coordinates of the points by substituting the value of θ into the equation for the ellipse.

4. What is the significance of the tangent of an ellipse being equal to 1?

When the tangent of an ellipse is equal to 1, it means that the slope of the tangent line at that point is equal to the slope of the line connecting the two focal points of the ellipse. This property is important in understanding the geometry of an ellipse and its relationship to its focal points.

5. Can you use calculus to find the points of an ellipse where the tangent equals 1?

Yes, you can use calculus to find the points of an ellipse where the tangent equals 1. By taking the derivative of the equation for the ellipse and setting it equal to 1, you can solve for the x and y coordinates of the points. This method is useful for finding the points of tangency for more complex equations of ellipses.

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