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Find the points on the curve $5x^2 - 6xy + 5y^2 = 4$ that are nearest the origin.

The first method I've tried is I've taken the derivative of the equation to optimize (Pythagorean Theorem) and also the function of the curve using implicit differentiation and plugged stuff in (the way I've done it so far). I got $y = x$ which I tried to plug into the function of the curve but didn't match the answers at the back of the textbook. I stared at it and figured I made a sign error somewhere since $-y = x$ worked but couldn't find it.

The second method I tried was solving for the function of the curve in terms of $y$ and tried both completing the square and using the quadratic formula to find $x$ which gave the same failed answer where $y$ subtracted to nonexistence when I plugged them into the function of the curve.

I don't know what lagrange multipliers are, which was the only solution on the web. I hope it could be done just with simple algebra (or calculus) since the textbook expects that (I think).

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  • $\begingroup$ Do you know what the curve looks like? $\endgroup$ – Calvin Lin May 28 '13 at 13:58
  • $\begingroup$ I assume it's an ellipse where the origin is somewhere inside of it. Graphing programs can't graph it and for some reason it never occurred to me to do it myself on paper. $\endgroup$ – user56833 May 28 '13 at 14:07
  • $\begingroup$ Check your equations again, $y=x$ is a solution (to the maximum) and $y=-x$ is a solution (to the minimum). You then need to check the second order conditions to determine which is a max and which is a min. $\endgroup$ – Calvin Lin May 28 '13 at 14:08
  • $\begingroup$ Yes, it is an ellipse. You can read my hint to understand why. $\endgroup$ – Calvin Lin May 28 '13 at 14:09
  • $\begingroup$ By the way you said something concerning; it seemed like you said that you solved for y and then plugged in back into your equation. If that is true, it is no wonder that the y subtracted into nonexistence, because if you substitute an equation into itself you always get the identity 0=0 $\endgroup$ – Ovi May 28 '13 at 16:38
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Hint: Using a change of coordinates $X=x+y$, $Y= x-y$, the problem becomes

$$ 5\frac{X^2+Y^2}{2} -6\frac{X^2-Y^2}{4} = 4,$$

or that

$$ X^2 + 4Y^2 = 4.$$

You should recognize this as an ellipse, with minor axis __ and major axis __.

Hence, the answer is __.


With regards to your differentiation solution:

$x^2 + y^2$ is minimized, when $2x + 2y \frac{dy}{dx} = 0$, or that $\frac{dy}{dx} = -\frac{y}{x}$.

In the original equation, we have $ 10x - 6y - 6x \frac{dy}{dx} + 10y \frac{dy}{dx}$, or that $\frac{dy}{dx} = \frac{6y-10x}{10y-6x} $.

Hence, solving for $-\frac{y}{x} = \frac{dy}{dx} = \frac{6y-10x}{10y-6x}$, we get $-y(10y-6x)=x(6y-10x)$, or that $-10y^2 +6xy = 6xy - 10x^2$, so $10(x^2-y^2)=0$. This has solution set $x=y, x=-y$.

You then need to check the second order condition, to see which (if any) gives your a global minimum.

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  • $\begingroup$ It's good to know that I was at least in the right path the first time. The last simplification 10(x^2 + y^2) = 0 never occurred to me and just simplified it to x = y. As for your hint, I'd have to look that up. $\endgroup$ – user56833 May 28 '13 at 14:56
  • $\begingroup$ Yes, it is a very common mistake for students to go from $x^2 = 1$ directly to $x=1$. You should factorize instead of 'taking square roots'. As for the hint, it has to do with taking a new rectilinear basis, which gets rid of the cross term (xy), and hence make it easier to understand the conic section. $\endgroup$ – Calvin Lin May 28 '13 at 15:03
  • $\begingroup$ Your transformation of the ellipse to rotated coordinates nicely avoids the ambiguity which arises in directly optimizing the distance function here. Because (in this problem) the ellipse has diagonal symmetry, implicit differentiation and Lagrange multipliers can't avoid the fact that the level curves for $ \ x^2 + y^2 \ $ are tangent to the ellipse along both the $ \ y = x \ $ and $ \ y = -x \ $ directions. $\endgroup$ – colormegone May 28 '13 at 17:05
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    $\begingroup$ @user56833 If you compare what Calvin Lin shows in the differentiation solution to what you found online using Lagrange multipliers, you'll see a good part of what the Lagrange method is doing (it is a worthwhile method to get to know). $\endgroup$ – colormegone May 28 '13 at 17:08
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Let $x=r\cos\theta$ and $y=r\sin\theta$. Then $$4=5x^2-6xy+5y^2=r^2-6r^2\cos\theta\sin\theta=r^2(5-3\sin 2\theta).$$ To minimize $r^2$, we maximize $5-3\sin 2\theta$. The maximum value of $5-3\sin 2\theta$ is $8$.

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  • $\begingroup$ This approach also nicely avoids the spurious (maximal-distance) solutions along the $ \ y = x \ $ direction that simple differentiation or "Lagrange" produces, as well as providing directly the square of the minimal distance. $\endgroup$ – colormegone May 28 '13 at 17:13
  • $\begingroup$ Can you please explain me why/how this method works? Moreover since x and y are the point lying on the curve so wouldn't the condition $x^2+y^2=r^2 $ need to be satisfied by every point on the curve? $\endgroup$ – Dhiraj Barnwal Apr 8 '16 at 5:53
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    $\begingroup$ @DhirajBarnwal: Here $r$ and $\theta$ are variables, just like $x$ and $y$ are. I just expressed the relationship between $x$ and $y$ as a relationship between $r$ and $\theta$, that is, found the polar equation of the curve . This polar equation is more useful for solving the minimal distance to the origin problem. Yes, $x^2+y^2=r^2$ holds identically, but $r$ is not constant. Alternately, one could use Lagrange multipliers on the rectangular coordinates version. More work. $\endgroup$ – André Nicolas Apr 8 '16 at 6:01
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You have to solve the following optimization problem:

Min $x^2+y^2$

Subject to:

$5x^2 -6xy+5y^2=4$

You could solve it using calculus as follows. Use the constraint to express $y$ in terms of $x$ and substitute that in the objective function. Then use the usual method of finding the optimal value by setting the $f'(x)=0$ and checking that $f''(x) >0$.

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  • $\begingroup$ I know. It's just that getting to the answer has been so far impossible. Getting the derivative of the function to optimize and the function of the curve then plugging that into the function to optimize doesn't match the answer at the back. That's what I've been doing so far. Had that been a 3-dimensional function, it would have been a lot easier. $\endgroup$ – user56833 May 28 '13 at 13:57

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