Find the point (or points) on the curve $y^2 = 4 + 3xy$ that are closest to the origin.

Question

I'm supposed to use Lagrangian Multipliers but I get stuck after getting the equations.

$f(x,y)=4+3xy-y^2$

$g(x,y)=x^2+y^2$

$3y=\lambda(2x)$

$3x-2y=\lambda(2y)$

I don't really know where to go from here as solving for $\lambda$ doesn't really give me anything useful.

Answer 1

Let $x^2+y^2=k>0$.

Thus, the following equation has solutions. $$y^2=\frac{4(x^2+y^2)}{k}+3xy$$ or $$\frac{4}{k}x^2+3xy+\left(\frac{4}{k}-1\right)y^2=0,$$ which gives $$9-\frac{16}{k}\left(\frac{4}{k}-1\right)\geq0$$ or $$k\geq\frac{8(\sqrt{10}-1)}{9}.$$ Id est, this point we obtain for $$k=\frac{8(\sqrt{10}-1)}{9},$$ $$x=-\frac{3y}{2\cdot\frac{4}{k}}$$ and $$y^2=4+3xy,$$ which gives two very ugly points:

$$\left(-\frac{2(\sqrt{10}-1)}{3\sqrt[4]{10}},\sqrt[4]{\frac{8}{5}}\right)$$ and $$\left(\frac{2(\sqrt{10}-1)}{3\sqrt[4]{10}},-\sqrt[4]{\frac{8}{5}}\right)$$

Answer 2

In your notation, you are supposed to be minimizing $g(x,y) = x^2 + y^2$ subject to $f(x,y) = y^2 - 3xy - 4 = 0$. So indeed, you must solve $f(x,y)=0$ and $\nabla g = \lambda \nabla f$.

So $\nabla g = (2x,2y)$ and $\nabla f = (-3y, 2y-3x)$ and you get $$ \begin{split} 2x &= \lambda (-3y) \\ 2y &= \lambda (2y-3x) \\ 4 &= y^2 - 3xy \end{split} $$ Can you now finish?

UPDATE

Here is one approach. Note that if $\lambda = 0$ then $x=0=y$ from the first two constraints, and the 3rd constraint is violated, hence $\lambda \ne 0$.

Similarly, $y \ne 0$ (otherwise 3rd constraint is violated directly) and $2y \ne 3x$, otherwise $y=0$.

Now divide the first 2 equations by each other (you can, since both sides of the second equation are non-zero). You get $$ \frac{x}{y} = \frac{3y}{3x-2y} $$ Cross-multiply the proportion to get $$ 3y^2 = x(3x-2y) = 3x^2 - 2xy $$ which we can simplify to $$ 0 = 3x^2 - 3y^2 + 2xy $$ Now you can solve the 3rd constraint for $x = 4/(y^2-3y)$ and plug it in here to get a quartic in $x$.

Answer 3

$$y^2=4+3xy$$ Use polar co-ordinates $x=r \cos \theta, y= r\sin \theta$. Then $$r^2\sin^2 \theta=4+3 r^2 \sin \theta \cos \theta \implies r^2(1-\cos 2\theta)a=4+3r^2\sin 2\theta$$ $$\implies r^2=\frac{4}{1- \cos2 \theta -3 \sin 2\theta}=\frac{4}{f(\theta)} \implies f(\theta)=1-\cos 2\theta-3 \sin 2\theta $$ $$\implies f'(\theta)=2\sin 2\theta-6\cos 2 \theta=0 \implies \tan 2\theta= 3,~ 2\theta=\frac{n\pi}{2}+\frac{1}{2}\tan^{-1} {3}$$ Let $n=1$, $\theta_1=\frac{\pi}{2}+\frac{1}{2}\tan^{-1}$ $$f''(\theta_1)=4\cos 2 \theta+ 12 \sin 2 \theta \implies f''(\theta_1)=- 4\sqrt{10}<0$$ So the periodic function will have max. as $f_{max}=f(\theta_1)=1+\sqrt{10}$ $$\implies r_{min}^2=\frac{4}{1+\sqrt{10}}\implies r_{min}=\frac{2}{3}\sqrt{\sqrt{10}-1}$$

Answer 4

If you swap the roles of $f$ and $g$, I think you get a much easier set of equations to deal with: $$2x = 3\lambda y \\ 2y = \lambda(3x-2y) \\ y^2=3xy+4.$$ From the first one, we have $x=\frac32\lambda y$, and substituting this into the second one produces $$2y = \frac12\lambda(9\lambda-4)y,$$ So either $y=0$ or $\lambda(9\lambda-4)=4$. The third equation of the system (the constraint) eliminates the first possibility; the second is a simple quadratic equation in $\lambda$. Solve that, and the rest of the solution proceeds by back-subtitution.