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

[ehild]

Now i want to solve in different methods and want to find the angle with which to rotate the axes like Ray said

Using polar coordinates might be the easiest method to find the extreme distances from the origin:

x = r cos(θ) ; y = r sin(θ) --> x² + y² = r² . Rewrite the equation of the curve also in terms of r and θ.
Find the extremes of r².
.

 

[epenguin]

Well I think it is a shame having pushed your calculations nearly to a conclusion, not to give the final bit and the conclusion.
I expect your rotation idea will give some instruction.
As you have made a good effort I think it is okay for me to give what looks to me like the approach needing as minmal calculation as possible

On the ellipse defined by
$$5x^2-6xy+5y^2=4~~~~~~~~~~(1a)$$
or
$$x^2+y^2-\frac{6}{5}xy=\frac{4}{5}~~~~~~~~~~~~(1b)$$
as convenient, we seek points where r² is minimum. I.e. it is necessary
$$\dfrac {d(r^2)}{dx}= 0 $$
which is
$$\dfrac {d}{dx}(x^2+y^2)= 0~~~~~~~~~~~~~~~~(2)$$
from which
$$\dfrac {dy}{dx}=-\dfrac {x}{y}~~~~~~~~~~~~~~~(3)$$
as you already noted in #1.Along the curve defined by (1)
$$\dfrac {dy}{dx}=\dfrac {d}{dx}\left( x^{2\cdot }+y^{2}-\dfrac {6}{5}xy\right)~~~~(4)$$

But by (2) the equation dy/dx = 0 for the sought extremal points simplifies to

$\dfrac {dy}{dx}=\dfrac {d}{dx}\left( \dfrac {-6}{5}xy\right) =0$ or just $~~~\dfrac {dy}{dx}=\dfrac {d}{dx}\left(xy\right)=0 ~~~~(5)$

(We could also say the curve (in this case an ellipse) Is cotangent to the circle at this closest point (see fig. in #29) I.e. has the same slope as our circle at this point.)
At the extremal points then from (5)
$$\dfrac {dy}{dx}=-\dfrac {y}{x}~~~~~~~~~~~(6)$$
Comparing (6) with (3) therefore, at these points

$x^2=y^2$, so $x=y$ or $x=-y$

Substituting these relations in (more conveniently) 1a you very easily get two values of $x^2$ and can quite easily conclude that the extremal points are (1, 1), (-1, -1) - those seen in #30 - and (-½, ½), (½, -½) seen in #(29).

 

[Ray Vickson]

Well I think it is a shame having pushed your calculations nearly to a conclusion, not to give the final bit and the conclusion.
I expect your rotation of idea will give some instruction.
As you have made a good effort I think it is okay for me to give what looks to me like the approach needing as minmal calculation as possible

On the ellipse defined by
$$5x^2-6xy+5y^2=4~~~~~~~~~~(1a)$$
---------------------------------------------------------------------------------

How about an approach that does not need calculus at all? Setting $x = r \cos \theta$ and $y = r \sin \theta$, the equation becomes
$$4 = 5 r^2 - 6 r^2 \sin \theta \; \cos \theta = r^2(5 - 3 \sin(2 \theta)), $$
or $r^2 = 4/(5 - 3 \sin (2 \theta)).$ For the first and second quadrants we have:
(1) Minimum $r^2$ ↔ maximum denominator ↔ $\sin (2 \theta)=-1$ ↔ $2 \theta$ = 270° ↔ $\theta$ = 135°, and $-x = y > 0$.
(2) Maximum $r^2$ ↔ minimum denominator ↔ $\sin(2 \theta)= 1$ ↔ $2 \theta$ = 90° ↔ $\theta$ = 45°, and $x = y > 0$.

 

[epenguin]

How about an approach that does not need calculus at all? Setting $x = r \cos \theta$ and $y = r \sin \theta$, the equation becomes
$$4 = 5 r^2 - 6 r^2 \sin \theta \; \cos \theta = r^2(5 - 3 \sin(2 \theta)), $$
or $r^2 = 4/(5 - 3 \sin (2 \theta)).$ For the first and second quadrants we have:
(1) Minimum $r^2$ ↔ maximum denominator ↔ $\sin (2 \theta)=-1$ ↔ $2 \theta$ = 270° ↔ $\theta$ = 135°, and $-x = y > 0$.
(2) Maximum $r^2$ ↔ minimum denominator ↔ $\sin(2 \theta)= 1$ ↔ $2 \theta$ = 90° ↔ $\theta$ = 45°, and $x = y > 0$.

Thank you for jogging me out of lazy calculus habit to reflect that it does not need to calculus to find that the slope of the tangent to a circle at point $(x, y)$ is $-\frac{x}{y}$, nor does it for finding that to change the shape of a rectangle of sides $x, y$ while keeping the area constant equal to $xy$, $\dfrac {\Delta y}{\Delta x}=\dfrac {-y}{x}$. The problem does not need calculus, nor trigonometry either.

 

[Karol]

Just substitute $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$ into your function $f(x,y) = 5(x^2+y^2) -6 xy$, to get a function $F(X,Y)$. Determine $\theta$ by requiring that there be no product term $X Y$ in the final $F$.

$$5(x\cos-y\sin)^2-6(x\cos-y\sin)(x\sin+y\cos)+5(x\sin+y\cos)^2=4$$
Only the xy term:
$$\sin^2=\cos^2~\rightarrow~\alpha=\pm 45^0$$
The new ellipse:
$$5(X^2+Y^2)=4~\rightarrow~5\left[ \left( \frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y \right)^2+\left( \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y \right)^2 \right]=4$$
$$s^2=X^2+Y^2=\left( \frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y \right)^2+\left( \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y \right)^2=X^2$$
$$s=\pm X~\rightarrow~s'=\pm 1$$
It's not logical. when x=0 the distance is 0?

 

[Ray Vickson]

$$5(x\cos-y\sin)^2-6(x\cos-y\sin)(x\sin+y\cos)+5(x\sin+y\cos)^2=4$$
Only the xy term:
$$\sin^2=\cos^2~\rightarrow~\alpha=\pm 45^0$$
The new ellipse:
$$5(X^2+Y^2)=4~\rightarrow~5\left[ \left( \frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y \right)^2+\left( \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y \right)^2 \right]=4$$
$$s^2=X^2+Y^2=\left( \frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y \right)^2+\left( \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y \right)^2=X^2$$
$$s=\pm X~\rightarrow~s'=\pm 1$$
It's not logical. when x=0 the distance is 0?

No, you have made a number of serious errors.
If $x = c X - sY$ and $y = sX+cY$, with $c = \cos \theta, s = \sin \theta$, then $5(x^2+y^2) = 5(X^2+Y^2)$ and $6 x y = cs(6X^2-6Y^2) +6XY(c^2-s^2)$.
The "$XY$" term is eliminated if $c = s = 1/\sqrt{2}$, giving $6xy = 3X^2 - 3Y^2$. Thus, $F = 5(x^2+y^2)-6xy$ $= 5X^2+5Y^2-3X^2+3Y^2$ $= 2X^2 + 8Y^2.$ The curve is $2X^2 + 8Y^2 = 4$.

 

[Karol]

$$2x^2+8y^2=4~\rightarrow~y^2=\frac{1}{2}-\frac{1}{4}x^2$$
$$s^2=x^2+y^2=x^2+\frac{1}{2}-\frac{1}{4}x^2=\frac{3}{4}x^2+\frac{1}{2}$$
$$s^2_{min}(x=0)=\frac{1}{4}~\rightarrow~s_{min}=\frac{1}{2}$$

One more approach:
Draw the picture.
apply $GM \leq AM$
$xy = (x^2 y^2 )^{\frac{1}{2}} \leq \gamma (x^2 + y^2)$
where $\gamma$ is some positive scalar we don't need to worry about.
Via two different domain restrictions, you get conditions for maximum distance $r^2$ and minimum distance $r^2$. (I.e. there are 4 quadrants in the graph, but you only need to worry about 2 of them, as symmetry then covers the other two.)

$$5(x^2+y^2)-6xy=4~\rightarrow~xy=\frac{5}{6}(x^2+y^2)-\frac{2}{3}$$
$$\gamma (x^2 + y^2)=\gamma s^2 \geq (x^2 y^2 )^{\frac{1}{2}} = xy=\frac{5}{6}(x^2+y^2)-\frac{2}{3}$$
$$\gamma s^2 \leq \frac{12}{5-6\gamma}$$
It doesn't lead to anything.
I tried an other thing:
$$xy=\sqrt{x^2y^2}=\sqrt{ \left[ \frac{5}{6}(x^2+y^2)-\frac{2}{3} \right ]}^2=\left[ \frac{25}{36}(x^2+y^2)^2-\frac{20}{18}(x^2+y^2)+\frac{4}{9} \right]^{1/2} \leq \gamma(x^2+y^2)$$
No where

 

[StoneTemplePython]

$$2x^2+8y^2=4~\rightarrow~y^2=\frac{1}{2}-\frac{1}{4}x^2$$
$$s^2=x^2+y^2=x^2+\frac{1}{2}-\frac{1}{4}x^2=\frac{3}{4}x^2+\frac{1}{2}$$
$$s^2_{min}(x=0)=\frac{1}{4}~\rightarrow~s_{min}=\frac{1}{2}$$$$5(x^2+y^2)-6xy=4~\rightarrow~xy=\frac{5}{6}(x^2+y^2)-\frac{2}{3}$$
$$\gamma (x^2 + y^2)=\gamma s^2 \geq (x^2 y^2 )^{\frac{1}{2}} = xy=\frac{5}{6}(x^2+y^2)-\frac{2}{3}$$
$$\gamma s^2 \leq \frac{12}{5-6\gamma}$$
It doesn't lead to anything.
I tried an other thing:
$$xy=\sqrt{x^2y^2}=\sqrt{ \left[ \frac{5}{6}(x^2+y^2)-\frac{2}{3} \right ]}^2=\left[ \frac{25}{36}(x^2+y^2)^2-\frac{20}{18}(x^2+y^2)+\frac{4}{9} \right]^{1/2} \leq \gamma(x^2+y^2)$$
No where

$GM \leq AM$ seems to be one of my favorite inequalities these days, so what I was thinking of was actually was

$$5x^2-6xy+5y^2=4~\rightarrow~x^2+y^2=\frac{4+6xy}{5}$$

your goal: minimize $x^2+y^2 = LHS$. This is an equality, so the Left Hand Side (LHS) is minimized if and only if the Right Hand Side (RHS) is.

$RHS = \frac{4+6xy}{5} = \frac{4}{5} + \frac{6}{5}xy = constant + \frac{6}{5}xy$

an additive constant is not directly important in optimization problems (another way to think about why they get mapped to zero by the derivative operator), so the RHS is minimized if and only if $\frac{6}{5}xy$ is minimized which is minimized if and only if $xy$ is minimized.

Draw the picture and split cases into two.

Top right and bottom left quadrants of your graph
$RHS = constant + \frac{6}{5}\big \vert xy \big \vert$

the scaling is positive, so $\frac{6}{5}\big \vert xy \big \vert \geq 0$

Top left and bottom right quadrants
$RHS = constant + \frac{3}{5}\Big(-1*\big \vert 2 xy \big \vert\Big)$

it's immediately apparent we want top left and bottom right quadrants, as we can scale the magnitude of $\big \vert 2 xy \big \vert$ by a negative number, and hence shrink the RHS.

Goal: maximize $\big \vert 2 xy \big \vert$. Apply $GM \leq AM$

$\big \vert 2 xy \big \vert = 2\big \vert x \big \vert \big \vert y \big \vert \leq \big \vert x\big \vert^2 + \big \vert y\big \vert^2 = x^2 + y^2$

with equality if and only if

$\big \vert x \big \vert= \big \vert y \big \vert$

and since we are in top left and bottom right quadrants, this means

$x = -y$
- - - -
edit: to be extra clear, the above tells us that for minimization we're dealing in top left or bottom right quadrant, and we have

$x^2 + y^2 = constant + \frac{3}{5}\Big(-1*\big \vert 2 xy \big \vert\Big) \geq constant - \frac{3}{5} (x^2 + y^2)$

add $\frac{3}{5} (x^2 + y^2)$ to each side
thus we have

$\frac{8}{5}\big(x^2 + y^2\big) \geq constant$

giving you the minimum value

$\big(x^2 + y^2\big) \geq \frac{5}{8}constant$

and from above, we know reaching this minimum occurs if and only if $x = -y$

- - - -

To be real clear, my approach is start by sketching the graph and looking at it. Then I look at the symbols involved -- in this case I see products on RHS and sums on LHS and I look for a way to go from products to sums. $GM \leq AM$ is one way to do this.

 

[Ray Vickson]

$$2x^2+8y^2=4~\rightarrow~y^2=\frac{1}{2}-\frac{1}{4}x^2$$
$$s^2=x^2+y^2=x^2+\frac{1}{2}-\frac{1}{4}x^2=\frac{3}{4}x^2+\frac{1}{2}$$
$$s^2_{min}(x=0)=\frac{1}{4}~\rightarrow~s_{min}=\frac{1}{2}$$

No! The original problem used $x,y$ and the rotated problem used $X,Y$. These are not at all the same, and should never, ever, be mixed up! I hope you are starting to learn that it is important to be more careful.

 

[epenguin]

As can be seen I didn’t get it immediately, maybe somebody has hinted it, but isn’t the easiest method the old quadratic inequalities tricks?

Re-write the equation

$$5x^2-6xy+5y^2=4$$
as
$$ 8\left( x^{2}+y^{2}\right) =4+3\left( x+y\right) ^{2}$$
Then the smallest possible value of the squared distance from origin $(x^2+y^2)$ is obtained when $x=-y$ and is ½. As $x^2=y^2$ at the points we easily obtain $x=\pm \dfrac {1}{2},y= \mp \dfrac {1}{2}$, giving the required distance $\sqrt {x^{2}+y^{2}}=\dfrac {1}{\sqrt {2}}$

Finding the points of maximum distance is left as an exercise for the reader.

 

[Karol]

No! The original problem used $x,y$ and the rotated problem used $X,Y$. These are not at all the same, and should never, ever, be mixed up! I hope you are starting to learn that it is important to be more careful.

$$2X^2 + 8Y^2 = 4~\rightarrow~Y^2=\frac{1}{2}-\frac{1}{4}X^2$$
$$s^2=X^2+Y^2=X^2+\frac{1}{2}-\frac{1}{4}X^2=\frac{3}{4}X^2+\frac{1}{2}$$
$$s^2_{min}(X=0)=\frac{1}{4}~\rightarrow~s_{min}=\frac{1}{2}$$
I just want to arrange the $~GM \leq AM~$ method.
$$GM \leq AM~\rightarrow~\frac{x+y}{2}\geqslant \sqrt{xy}~\rightarrow~\frac{x^2+y^2}{2}\geqslant \sqrt{x^2 y^2}=xy~\rightarrow~2xy\leqslant x^2+y^2$$
$$5(x^2+y^2)-6xy=4~\rightarrow~x^2+y^2=\frac{4}{5}-\frac{3}{5}\vert 2xy \vert \geqslant \frac{4}{5}-\frac{3}{5}(x^2y^2)$$
$$\frac{8}{5}(x^2y^2)\geqslant \frac{4}{5}~\rightarrow~x^2+y^2 \geqslant \frac{1}{2}$$
I will review now my original solution.

I don't know the Lagrange method and i am sure it's not to be used, since this chapter is the third in the book, and the book started from "scratch", it taught systematically, and it didn't teach it.
$$s=x^2+y^2~\rightarrow~s'=2x+2yy'$$
I insert $~y'=-\frac{x}{y}~$: s'=0
So the distance is a constant and the equation is a circle.

$$5(x^2+y^2)-6xy=4~\rightarrow~10x-6(yy'+x)+10yy'=0~\rightarrow~y'=-\frac{x}{y}$$
Why isn't that correct? i know now it's an ellipse but what is wrong with the derivative of the distance?

 

[Ray Vickson]

$$5(x^2+y^2)-6xy=4~\rightarrow~10x-6(yy'+x)+10yy'=0~\rightarrow~y'=-\frac{x}{y}$$
Why isn't that correct? i know now it's an ellipse but what is wrong with the derivative of the distance?

$$5(x^2+y^2) - 6xy=4 \: \Rightarrow 10 x + 10 y y' - 6 y - 6 xy' = 0 \; \Rightarrow y' = \frac{5x-3y}{3x-5y}.$$

 

[Karol]

$$y' = \frac{5x-3y}{3x-5y},~~y'=0~\rightarrow~5x-3y=0~\rightarrow~y=\frac{5}{3}x$$
$$5(x^2+y^2)-6xy=4~\rightarrow~x^2=-\frac{18}{5}$$

 

[epenguin]

$y’=0$ is not the point you are looking for - it is in no way an answer to the question. Look at pictures of the ellipse, e.g. #28 to see what points $y’=0$ corresponds to.

 

[Ray Vickson]

$$y' = \frac{5x-3y}{3x-5y},~~y'=0~\rightarrow~5x-3y=0~\rightarrow~y=\frac{5}{3}x$$
$$5(x^2+y^2)-6xy=4~\rightarrow~x^2=-\frac{18}{5}$$

No, again. There are two expressions for $y'$:
(1) From the optimality condition:
$$0 = \frac{d}{dx} (x^2+y^2) = 2 x + 2 y y' \Rightarrow y' = -\frac{x}{y}$$
(2) From the constraint:
$$ 0 = \frac{d}{dx} (5 x^2 + 5y^2 - 6 xy) = 10 x + 10 y y' - 6y - 6 xy' \Rightarrow y' = \frac{5x-3y}{3x-5y}$$
How would you use those two formulas for $y'$ to get further along towards a solution?

 

[Karol]

$$5(x^2+y^2) - 6xy=4 \: \Rightarrow \: s^2=x^2+y^2=\frac{4}{5}+\frac{6}{5}xy$$
$$[s^2]'=\frac{6}{5}(y+xy')=\frac{6}{5}\left[ y+\frac{5x-3y}{3x-5y} \right]$$
I can't express y(x)

 

[Karol]

$$\frac{5x-3y}{3x-5y}=-\frac{x}{y}~\rightarrow~x^2=y^2$$
$$x=y,~x=-y,~-x=y,~-x=-y$$
The 4 solutions are contained in only two: $~x=y,~x=-y~$. now to try each one in the original formula:
$$x=y:~5x^2-6x^2+5x^2=4~\rightarrow~x=\pm 1$$
$$x=-y:~5x^2+6x^2+5x^2=4~\rightarrow~x=\pm \frac{1}{2}$$
The first is max. distance and the second min.

 

[epenguin]

You got it! To be pedantic a slight mistake in your last sentence which might lose you marks, not quite the answer to the question asked, but we know what you mean - you are probably exhausted by now.

More important, the job is not finished! You need to apply a Polya principle:
“4 look back on your work. How could it be better?“

The last posts here are unnecessarily complicated. The calculation of $\frac{d}{dx} (5 x^2 + 5y^2 - 6 xy)$ is carried through in full without exploiting the fact that $\frac{d}{dx} (x^2+y^2) = 0$ (at the point of interest) so you only need to find the condition for $\frac{d}{dx}(xy)=0$ (which is essentially what I have done in #37). Since #15 I have been trying to prod you to recognise and exploit the fact that the distance function $(x^2+y^2)$ Is also part of the equation of the curve.

The ‘Polya principles’ to which I often refer in this site are contained in the slim, cheap and popular book “How to Solve It“ by George Polya https://en.wikipedia.org/wiki/How_to_Solve_It which is worth at least a dozen or two maths lessons for giving you mathematical muscle. It is worth buying and not just reading but keeping by. Because when you read it you may think ah that’s fairly obvious – but then you forget it and don’t apply at it when needed. One tip in it is to ‘draw a picture’, I have the impression you didn’t at first and it would have helped see where you were going. As I say one forgets. A tip that has got me unstuck sometimes is ”are you using all the information available?“ - But then in the present case I had lost sight of another one: “Have you seen anything like it before?”. And sure I must have done lots of exercises at school and elsewhere on inequalities (and maxima and minima are inequalities) using squares whose minimum value is zero. And so probably have you, so you too did have the necessary background in theory. After a bit this came out of the back of my mind and I arrived at #45 which looks to be the most efficient solution possible. Which leads me to my final heuristic point.

You were asked at the start of the thread #2 whether you knew the general method for constrained extrema. General methods are surely the biggest most powerful thing in maths. But when anything can be solved by a less general method, that will usually be more efficient because it is using more information about the specifics of the problem.

 

[WWGD]

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 :

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

 

[WWGD]

You got it! To be pedantic a slight mistake in your last sentence which might lose you marks, not quite the answer to the question asked, but we know what you mean - you are probably exhausted by now.

More important, the job is not finished! You need to apply a Polya principle:
“4 look back on your work. How could it be better?“

You were asked at the start of the thread #2 whether you knew the general method for constrained extrema. General methods are surely the biggest most powerful thing in maths. But when anything can be solved by a less general method, that will usually be more efficient because it is using more information about the specifics of the problem.

Re first point: always good to keep tweaking.

RE second point, problem is that it takes a good amount of general mental computing resources and time to come up with customized methods. This is why, I would say, automation through standardization is so often used.

 

[StoneTemplePython]

More important, the job is not finished! You need to apply a Polya principle:
“4 look back on your work. How could it be better?“

+1 for a great Polya reference. I am big fan of trying to solve a problem in many different ways as well.

There's something of an art to figuring out how to tailor your approach to the specifics of the problem. (I think people call this mathematical creativity.)

 

[Karol]

(1) Solve the curve equation for y in terms of x; that has two roots, one for the upper part of the curve, and one for the lower part. For each part you can substitute in your formula for y = y(x), to get a squared distance $s = x^2 + y^2$ that is a function of $x$ alone, then use standard calculus methods to find maxima and minima for each of the two parts.

$$5x^2-6xy+5y^2=4~\rightarrow~5y^2-(6x)y+(5x^2-4)=0$$
$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}=\frac{3x\pm 2\sqrt{5-4x^2}}{5}$$
I choose the + first:
$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2=x^2+\frac{1}{25}[12x\sqrt{5-4x^2}+20-7x^2]$$
$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]$$
$$s'=0~\rightarrow~x=\pm 1.1$$

 

[Karol]

I posted a few days ago but nobody addressed

 

[epenguin]

Why I could almost have said the same thing myself a few times.
In your case just a few more words would have done something for comprehending where you are at (I mentioned ‘narrative’ before). Readers are left uncertain whether you mean at the end really one point one, or whether you meant just 1, which if I remember does correspond to a maximum of $x$.You need to say (if this is what you mean) this is wrong, but say how you know it’s wrong. Otherwise people just only see a blank mass of calculation.

And if they get that much into it they may well wonder how you got x from s’ = 0 - It looks like having to go through the fourth power palaver or worse again. I don’t suggest you set all the detail out but some words to indicate what you did would not be out of place.I wonder how many pages you have written? Without any words I think you won’t yourself know what you’ve done or why when you come back to it in six months.

I am not very good at arithmetic, but it seems to be on the second line you made a maltranscription or something and under the square root you should have (5² - 4²x²) - the squarings have slipped away.

 

[ehild]

I posted a few days ago but nobody addressed

Karol, you knew your result was wrong. Why did you post it?
How did you get that wrong result from your last correct formula?

 

[Karol]

(1) Solve the curve equation for y in terms of x; that has two roots, one for the upper part of the curve, and one for the lower part. For each part you can substitute in your formula for y = y(x), to get a squared distance $s = x^2 + y^2$ that is a function of $x$ alone, then use standard calculus methods to find maxima and minima for each of the two parts.

$$5x^2-6xy+5y^2=4~\rightarrow~5y^2-(6x)y+(5x^2-4)=0$$
$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}=\frac{3x\pm 2\sqrt{5-4x^2}}{5}$$
I choose to start with the positive root:
$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2=x^2+\frac{1}{25}[12x\sqrt{5-4x^2}+20-7x^2]$$
$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]=\frac{60-96x^2-300x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
$$s'=0~\rightarrow~60-96x^2-300x\sqrt{5-4x^2}=0$$
I can't solve

 

[epenguin]

Sorry I have already explained how your equation of line 1 is drastically simplified for s’ = 0.
But maybe you want to show that without that you can arrive at the results all the same. However you have taken no account of my observation that there seems to be an error in line 2 - you repeat what you wrote before. So please deal with this, if it is the case tell me I’m wrong.

I think that at the end you will get an equation of the same general kind as your final one you say you can’t solve. But an equation of that kind can be solved algebraically – say the equation is your expression = 0, you just take last term of the left hand to the right hand side of the equation then square both sides and you get an equation which is a quadratic in the variable x².

 

[Ray Vickson]

$$5x^2-6xy+5y^2=4~\rightarrow~5y^2-(6x)y+(5x^2-4)=0$$
$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}=\frac{3x\pm 2\sqrt{5-4x^2}}{5}$$
I choose to start with the positive root:
$$s=x^2+y^2~\rightarrow~s_{+y}=x^2+\left[ \frac{3x+2\sqrt{5-4x^2}}{5} \right]^2=x^2+\frac{1}{25}[12x\sqrt{5-4x^2}+20-7x^2]$$
$$s'=2x+\frac{1}{25}\left[ 12 \left( \sqrt{5-4x^2}-\frac{4x^2}{\sqrt{5-4x^2}} \right)-14x \right]=\frac{60-96x^2-300x\sqrt{5-4x^2}}{25\sqrt{5-4x^2}}$$
$$s'=0~\rightarrow~60-96x^2-300x\sqrt{5-4x^2}=0$$
I can't solve

Write it as $300x \sqrt{5-4x^2} = 60 - 96 x^2$. Now square both sides and you will get an equation that is quadratic in the variable $w = x^2$.

 

[Karol]

However you have taken no account of my observation that there seems to be an error in line 2 - you repeat what you wrote before.So please deal with this, if it is the case tell me I’m wrong.

$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}$$
Under the square root (i forgot how it's called):
$$36x^2-20(5x^2-4)=80-64x^2=16(5-4x^2)$$

 

[epenguin]

$$y=\frac{6x\pm\sqrt{36x^2-20(5x^2-4)}}{10}$$
Under the square root (i forgot how it's called):
$$36x^2-20(5x^2-4)=80-64x^2=16(5-4x^2)$$

Okay and I had been through that several times – I said I was not good at arithmetic.

So now try and solve that in the way that I and Ray Vickson have indicated .

 

[Karol]

Write it as $300x \sqrt{5-4x^2} = 60 - 96 x^2$. Now square both sides and you will get an equation that is quadratic in the variable $w = x^2$.

$$300x \sqrt{5-4x^2} = 60 - 96 x^2~\rightarrow~90000(5-4x^2)=3600-11520x^2+9216x^4$$
$$625x^2(5-4x^2)=25-80x^2+64x^4$$
$$2564x^4-3205x^2+25=0~\rightarrow~2564a^2-3205a+25=0$$
$$\frac{3205\pm\sqrt{10015625}}{5128}$$
The root isn't round, so i didn't calculate till the end. i think it doesn't give any correct x.

 

[epenguin]

In

$$s'=0~\rightarrow~60-96x^2-300x\sqrt{5-4x^2}=0$$

it is helpful to remove numerical factors of the whole expression which operation do not change the roots.

 

[Karol]

Sorry I have already explained how your equation of a line one is drastically simplified for s’ = 0

Can you please direct me to the post number you explained that, epenguin? there are already many posts and i intend to go through all of them

 

[Karol]

In
it is helpful to remove numerical factors of the whole expression which operation do not change the roots.

$$300x \sqrt{5-4x^2} = 60 - 96 x^2~\rightarrow~25x\sqrt{5-4x^2}=5-8x^2$$
$$625x^2(5-4x^2)=400-1280x^2+1024x^4$$
$$3524x^4-4405x^2+400=0$$
$$\frac{4405\pm\sqrt{13765625}}{7048}$$
No exact root

 

[Ray Vickson]

$$300x \sqrt{5-4x^2} = 60 - 96 x^2~\rightarrow~25x\sqrt{5-4x^2}=5-8x^2$$
$$625x^2(5-4x^2)=400-1280x^2+1024x^4$$
$$3524x^4-4405x^2+400=0$$
$$\frac{4405\pm\sqrt{13765625}}{7048}$$
No exact root

So, you MUST HAVE made an error somewhere, because you already know the solution. Why do you keep posting incorrect results? Just go back and fix your errors---you don't need us to tell you that.

I quit.