Find the Points on an Ellipse Furthest Away From (1,0)

It makes life so much easier when trying to find the OP in longer threads.In summary, to find the points on the ellipse 4x2+y2=4 that are furthest from the point (1,0) on the ellipse, you can use the distance formula d=√[(x2-x1)2+(y2-y1)2] and set it equal to 0 to find the maxima points. Alternatively, you can take the derivative of the function d^2 with respect to x and set it equal to 0, as the maximum power of x in d^2 is 2. The maximum distance occurs at x=-1/3.
  • #1
Drakkith
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Homework Statement


Find the points on the ellipse 4x2+y2=4 that are furthest from the point (1,0) on the ellipse.

Homework Equations


Ellipse: y=±√(4-4x2)
Distance Formula: d=√[(x2-x1)2+(y2-y1)2]

The Attempt at a Solution


The distance from (1,0) for any point on the ellipse should be d=√[(x-1)2+(±√(4-4x2)2].
That simplifies to d=√[(x-1)2+4-4x2]. The ± in front of √(4-4x2) just turns positive because the final value is squared anyways, right?

However, when I graph the distance formula it doesn't match up with my ellipse. According to my solution manual the maximum distance occurs at x=-1/3, but my distance formula graph just keeps increasing. The manual squares d to give s=d2 and then graphs s, which does give a maximum value for s at x=-1/3, but I don't understand why they do this and how it works.

Edit: Added negative sign to x=-1/3.
 
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  • #2
Drakkith said:

Homework Statement


Find the points on the ellipse 4x2+y2=4 that are furthest from the point (1,0) on the ellipse.

Homework Equations


Ellipse: y=±√(4-4x2)
Distance Formula: d=√[(x2-x1)2+(y2-y1)2]

The Attempt at a Solution


The distance from (1,0) for any point on the ellipse should be d=√[(x-1)2+(±√(4-4x2)2].
That simplifies to d=√[(x-1)2+4-4x2]. The ± in front of √(4-4x2) just turns positive because the final value is squared anyways, right?

However, when I graph the distance formula it doesn't match up with my ellipse. According to my solution manual the maximum distance occurs at x=1/3, but my distance formula graph just keeps increasing. The manual squares d to give s=d2 and then graphs s, which does give a maximum value for s at x=1/3, but I don't understand why they do this and how it works.

You should show your work, because I am in agreement with your solution manual. Presumably, you have made an error somewhere.
 
  • #3
Ray Vickson said:
You should show your work,

I thought I did so. What else would you like to see?
 
  • #4
I'm not in agreement with your solution manual. I find the maxima occur at ##x=-1/3##. If you plot the ellipse, you should be able to see x=+1/3 doesn't make sense either.
 
  • #5
Whoops, I forgot to put the negative sign in front of the 1/3. You are correct, Vela, it is -1/3.
 
  • #6
The only thing I can think of is that you're not plotting the right function. What you wrote in the OP is fine. Whether you plot ##d## or ##d^2##, you should see a maximum at x=-1/3.
 
  • #7
Another approach that I find much simpler is to take the derivative of the function ##d^2## w.r.t ##x## and let that equal to 0 to find the maxima points.
since the maximum power of ##x## in ##d^2## is 2, taking the derivative will produce a linear equation and you should have your desired solution.
[EDIT:- @Drakkith you being a mentor why does your profile have the label STARTER on it?]
 
  • #8
vela said:
The only thing I can think of is that you're not plotting the right function. What you wrote in the OP is fine. Whether you plot ##d## or ##d^2##, you should see a maximum at x=-1/3.

Dang it, you are correct. I put 4-4X instead of 4-4X2 into my function. Now it has a maximum at x=-1/3.
I swear... I checked my work about 5 times and still missed that. Thanks, Vela.

certainly said:
Another approach that I find much simpler is to take the derivative of the function ##d^2## w.r.t ##x## and let that equal to 0 to find the maxima points.
since the maximum power of ##x## in ##d^2## is 2, taking the derivative will produce a linear equation and you should have your desired solution.

Thanks. I'll try that. I suppose it would be a good way to double check my answer.
[EDIT:- @Drakkith you being a mentor why does your profile have the label STARTER on it?]

Because I started this thread.
 
  • #9
Drakkith said:
Because I started this thread.
Oh I see... I always thought a starter was someone who was had just created his PF account a week earlier or something like that...
 
  • #10
certainly said:
Oh I see... I always thought a starter was someone who was had just created his PF account a week earlier or something like that...

Nope. It's just an easy way to see who started the thread. I'm really happy Greg added it.
 

1. What is an ellipse?

An ellipse is a closed curve formed by the intersection of a cone and a plane, where the plane cuts through the cone at an angle. It is a type of conic section and can be seen as a stretched or flattened circle.

2. How do you find the points on an ellipse that are farthest away from a given point?

To find the points on an ellipse that are furthest away from a given point, you can use the distance formula. First, find the equation of the ellipse in standard form. Then, substitute the coordinates of the given point into the distance formula to find the distance between the given point and any point on the ellipse. The points that have the largest distance from the given point will be the points on the ellipse that are furthest away.

3. What is the significance of the point (1,0) in this problem?

The point (1,0) represents the focus of the ellipse, which is a fixed point inside the ellipse that is used to construct the shape. In this problem, it is the point from which we want to find the points on the ellipse that are furthest away.

4. Can you use this method to find the points on any ellipse?

Yes, this method can be used to find the points on any ellipse. However, the given point and the ellipse must be in the same plane for this method to work. If the given point is not in the same plane as the ellipse, a different method will need to be used.

5. Are there any limitations to using this method?

One limitation of this method is that it may not give the most accurate results for ellipses with a high eccentricity (flattening). In these cases, it may be better to use a numerical or graphical method to find the points on the ellipse that are furthest away from the given point.

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