Graph drawing—Finding the points on a curve that are nearest to the origin

[Karol]

I know how to obtain $~\dfrac {dy}{dx}=-\dfrac {x}{y}~$, but i ask something else.
You used [r²]'=0 twice by taking into consideration that the distance is part of the original equation. it indeed simplified.
But isn't it too much to use the same consideration, same argument twice in the same process? it is a general question, not related only to this question.
The second time you used [r²]'=0 is at the end of #37 at:

Comparing (6) with (3) therefore, at these points

 

[Karol]

Just curious. How about finding the solution like this:
Find a circle C0 centered at the origin with radius r which intersects the ellipse at precisely one point? The value of r is then the shortest distance?

EDIT: $x^2+y^2=r^2 . 5x^2+5Y^2 -6XY=4 =5r^2-6x(\sqrt{r^2-x^2})=4$. Square both sides and find values of $r$ so that there is just one solution. $5r^2-4 =6x(\sqrt{ r^2-x^2} \rightarrow 25r^4-10r^2+16=36x^2(r^2-x^2) \rightarrow -36x^4 +36x^2r^2 -25r^4 +10r^2 -16 =0$ A bi-quadratic Set $r^2=s$ then we get :

$-36x^4 +36x^2s-25s^2 -10s-16=0. -25s^2-(10 -36x^2)s+ 36x^4-16=0 . s=10-36x^2 \pm \sqrt{}$ . .

$$\frac{10-36x^2\pm\sqrt{(10-36x^2)^2+100(36x^4-16)}}{-50}=\frac{10-36x^2\pm 2\sqrt{3}\sqrt{408x^4-60x^2-125}}{-50}$$

 

[WWGD]

$$\frac{10-36x^2\pm\sqrt{(10-36x^2)^2+100(36x^4-16)}}{-50}=\frac{10-36x^2\pm 2\sqrt{3}\sqrt{408x^4-60x^2-125}}{-50}$$

Sure, thanks, and sorrry for leaving without completing. Doesn't this work?

 

[Karol]

It doesn't work since the expression under the root doesn't give precise value.
$$s=\frac{10-36x^2\pm 2\sqrt{3}\sqrt{408x^4-60x^2-125}}{-50}$$
S is positive and because the denominator is negative (-50) the nominator must be positive.
I don't know to continue from here. under the root i denote $~x^2=a$:
$$408a^2-60a-125=0~\rightarrow~a=\frac{60\pm\sqrt{207,600}}{816}$$
I can't continue

 

[ehild]

Just curious. How about finding the solution like this:
Find a circle C0 centered at the origin with radius r which intersects the ellipse at precisely one point? The value of r is then the shortest distance?

EDIT: $x^2+y^2=r^2 . 5x^2+5Y^2 -6XY=4 =5r^2-6x(\sqrt{r^2-x^2})=4$. Square both sides and find values of $r$ so that there is just one solution. $5r^2-4 =6x(\sqrt{ r^2-x^2} \rightarrow 25r^4-10r^2+16=36x^2(r^2-x^2) \rightarrow -36x^4 +36x^2r^2 -25r^4 +10r^2 -16 =0$
.

The line highlighted by red is wrong. It should be
$5r^2-4 =6x(\sqrt{ r^2-x^2} \rightarrow 25r^4-40r^2+16=36x^2(r^2-x^2)$
But it is better to keep $(5r^2-4)^2$ un-expanded.

 

[Karol]

$$\left[ \frac{5r^2-4}{6} \right]^2=x^2(r^2-x^2)$$
Any value that solves is good, +x and -x are always answers.
I can't solve

 

[Dick]

Why not? Do you want to solve for $x$? It's a quadratic equation in the variable $x^2$.

 

[Karol]

$$x^2+y^2=r^2,~~5x^2+5Y^2 -6xy=4~\rightarrow~5r^2-4 =6x(\sqrt{ r^2-x^2})$$
$$\left[ \frac{5r^2-4}{6} \right]^2=x^2(r^2-x^2)~\rightarrow~\left[ \frac{5r^2-4}{6} \right]^2=r^2a-a^2$$
$$a^2-r^2a+\left[ \frac{5r^2-4}{6} \right]^2=0$$
If the root will be 0 there will be only one solution for a=x²
$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4-5r^2+2}}{2}$$
Under the root:
$$-2r^4-5r^2+2=0~\rightarrow~\frac{5\pm\sqrt{25+16}}{-4}$$
Cannot equal this to 0

 

[ehild]

$$x^2+y^2=r^2,~~5x^2+5Y^2 -6xy=4~\rightarrow~5r^2-4 =6x(\sqrt{ r^2-x^2})$$
$$\left[ \frac{5r^2-4}{6} \right]^2=x^2(r^2-x^2)~\rightarrow~\left[ \frac{5r^2-4}{6} \right]^2=r^2a-a^2$$
$$a^2-r^2a+\left[ \frac{5r^2-4}{6} \right]^2=0$$
If the root will be 0 there will be only one solution for a=x²
$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4-5r^2+2}}{2}$$

Check the signs.

 

[Karol]

$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4+5r^2-2}}{2}$$
Under the root, $~r^2=g~$:
$$-2g^2+5g-2=0~\rightarrow~g_1=2,~g_2=\frac{1}{2}$$
The shortest distance is $~g_2=r^2=\frac{1}{2}$
Back to the original equation $~5(x^2+y^2)-6xy=4~\rightarrow~5r^2-4=6xy~~\rightarrow~xy=-\frac{1}{4}$
I insert $~y=\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]~$:
$$xy=-\frac{1}{4}~\rightarrow~x\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{1}{4}$$
$$~\rightarrow~-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})$$
If i square both sides, i don't get rid of the root

 

[Dick]

$$\frac{r^2\pm\sqrt{r^4-4\left[ \frac{5r^2-4}{6} \right]^2}}{2}=\frac{r^2\pm\frac{2\sqrt{2}}{3}\sqrt{-2r^4+5r^2-2}}{2}$$
Under the root, $~r^2=g~$:
$$-2g^2+5g-2=0~\rightarrow~g_1=2,~g_2=\frac{1}{2}$$
The shortest distance is $~g_2=r^2=\frac{1}{2}$
Back to the original equation $~5(x^2+y^2)-6xy=4~\rightarrow~5r^2-4=6xy~~\rightarrow~xy=-\frac{1}{4}$
I insert $~y=\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]~$:
$$xy=-\frac{1}{4}~\rightarrow~x\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]=-\frac{1}{4}$$
$$~\rightarrow~-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})$$
If i square both sides, i don't get rid of the root

Don't square both sides until you have the square root isolated by itself.

 

[Karol]

$$-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})~\rightarrow~-\frac{5}{24x^3}=\pm\sqrt{5-4x^2}$$
After isolating the root and squaring both sides:
$$2304\cdot x^8-2880\cdot x^6+25=0,~~a=x^2~\rightarrow~2304\cdot a^4-2880\cdot a^3+25=0$$
And now what?

 

[tnich]

$$-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})~\rightarrow~-\frac{5}{24x^3}=\pm\sqrt{5-4x^2}$$
And now what?

This part is not correct.
$$x(3x\pm 2\sqrt{5-4x^2})\neq \pm 6x^3\sqrt{5-4x^2})$$

 

[Karol]

$$-\frac{5}{4}=x(3x\pm 2\sqrt{5-4x^2})~\rightarrow~-\frac{5+12x^2}{8x}=\pm\sqrt{5-4x^2}$$
After squaring:
$$400x^4-200x^2+25=0~\rightarrow~16x^4-8x^2+1=0~\rightarrow~16a^2-8a+1=0$$
$$~\rightarrow~\left[ a-\frac{1}{4} \right]^2=0~\rightarrow~x^2=\frac{1}{4},~x=\pm\frac{1}{2}$$

 

[tnich]

$$~x=\pm\frac{1}{2}$$

You are on the right track. Now all you have to do is figure out the corresponding values of y. Be careful with signs.
Once you have done that, try @ehild's suggestion and convert to polar coordinates. That is a much simpler way to solve this problem.

 

[epenguin]

You got there. Actually you got the answer (r) to the question essentially in #130. However, for completeness it is necessary to know the corresponding x and y. You have got x² and so the two x, and from x² and r² you can now get y². Then you can get y, however you previously saw that not every combination of and x and y that you get is allowable.
If that previous conclusion depended on a calculus argument, and here you want to do an algebraic argument, then you need to complete this.

You have been doing things a long way around in my opinion, but I hope you have learned some things.
For example I hope as soon as you saw

$$-2g^2+5g-2=0$$

in #130, you realized that one solution would be the reciprocal of the other.This happens when coefficients of a polynomial equation equidistant from the ends are equal. Called reciprocal equations their solution can be reduced to solving equations of half the degree plus a quadratic. In your case you only have a quadratic anyway, I don't think there is any saving.

The problem which was previously solved by calculus here you are solving by an algebraic argument. That depended on saying that two roots of an equation have to be equal at the shortest distance. I used the same argument in #45, or at least it could be worded that way. I think it has to be admitted that that was a whole lot faster and easier. Probably on analysis your argument could be shown to be equivalent.

To complete this learning and take on board this other point of view, I again invite you to use the approach of #45 to get the maximum distance. If you do not do this by about Monday I will do it here.

 

[Karol]

You have got x² and so the two x, and from x² and r² you can now get y²

Why do i get y² and not y? the original equation is $~5r^2-6xy=4~$ and from that, by inserting r² and x i can get directly the y and both values of y will be valid, because you said:

however you previously saw that not every combination of and x and y that you get is allowable.

 

[epenguin]

Ah yes if you do it that way it comes OK without ambiguity - I was thinking you might go and use r² and x² .

 

[Karol]

How do you do this kind of hyperlink:

 

[tnich]

How do you do this kind of hyperlink:
View attachment 220470

Just type "@" and the user name.

 

[Karol]

I thank you all for your patience:

Dick
Ray
WWGD
ehild
Phyton
epenguin
Nidum
Chestermiller
Mark44
tnich

It was a long ride

 

[epenguin]

It was a long ride

Though a short distance.
You took some long routes around where you saw some hopefully memorable scenery, like rotating axes, reciprocal equations amongst others.
Every method to find this shortest distance was trying to tell you at the same time the greatest distance, but this was always ignored. In order to fix the method of quadratic inequalities (#45) in your mind so that you might draw on it if you meet a suitable problem, I urged you find the largest distance using it. But as this hasn't been done I can't hold back from it .

Re-write the equation of your curve $5x^2-6xy+5y^2=4$ as
$$~~~~~~~~~~~~ 2(x^2+y^2) =4-3(x-y)^2$$
Then maximum of r^2 is 2 and the maximum of r is √2, which occurs when the square on the right is zero: there $x=y=±1$.

 

[Karol]

Then maximum of r^2 is 2

Why?

 

[Ray Vickson]

Why?

This has been explained to you many, many times already; just look at #4, #12, #38.

 

[Karol]

I didn't know you can mix methods. i wouldn't have gotten to that

 

[epenguin]

Why?

Because any square must be ≥ 0, now can the RHS of the expression in #142 be greater than 4? Similar principle was involved in #45.

 

[Ray Vickson]

I didn't know you can mix methods. i wouldn't have gotten to that

What dicatator told you that you cannot mix methods? Most good problem solvers mix methods all the time; learning to do that effectively should be part of any math education.

 

[Karol]

It was a complex problem for me and i still feel unsure.
I thank you all again
This site and your attitude are invaluable for me

 

[epenguin]

It was far more satisfactory to me to interact with a student who saw the thing through to the end, told us all his results and reasoning including misunderstanding, than one who does not do these things. Or even who tells you 'thanks I have solved it' but you wonder if he has as well as he thinks.