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

1. Apr 24, 2015

### Drakkith

Staff Emeritus
1. The problem statement, all variables and given/known data
Find the points on the ellipse 4x2+y2=4 that are furthest from the point (1,0) on the ellipse.

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

3. 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.

Last edited: Apr 24, 2015
2. Apr 24, 2015

### Ray Vickson

You should show your work, because I am in agreement with your solution manual. Presumably, you have made an error somewhere.

3. Apr 24, 2015

### Drakkith

Staff Emeritus
I thought I did so. What else would you like to see?

4. Apr 24, 2015

### vela

Staff Emeritus
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. Apr 24, 2015

### Drakkith

Staff Emeritus
Whoops, I forgot to put the negative sign in front of the 1/3. You are correct, Vela, it is -1/3.

6. Apr 24, 2015

### vela

Staff Emeritus
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. Apr 24, 2015

### certainly

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. Apr 24, 2015

### Drakkith

Staff Emeritus
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.

Thanks. I'll try that. I suppose it would be a good way to double check my answer.

9. Apr 24, 2015

### certainly

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. Apr 24, 2015

### Drakkith

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