0
$\begingroup$

I am stuck with the following problem :

Using Lagrange's method find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$.

My try: To find the min. value of $$r^2=x^2+y^2$$,where $r^2$ is the shortest distance from $(0,0)$ to any point $(x,y)$ of the hyperbola $x^2+8xy+7y^2=225$.

We construct the Lagrangian function $$\mathcal{L}(x,y,\lambda)=x^2+y^2+\lambda (x^2+8xy+7y^2-225)$$,where $\lambda$ is undetermined multiplier.

Now, The stationary points are given by : $\mathcal{L_x}=0$ and $\mathcal{L_y}=0$.

Now, $\mathcal{L_x}=0 \implies 1=-\lambda(1+4a) \tag{1}$ and $\mathcal{L_y}=0 \implies a=-\lambda(4+7a) \tag{2}$ where $a=\frac yx$.

Now, from $(1),(2)$, we get $a=2,-\frac 12$.

For $a=2$, we get from $(1), \lambda =-\frac 19$.

Now, $a=2 \implies \frac yx=2 \implies \frac y2=\frac x1 =k( \neq 0\,\,\text{say})$.

Now, I am stuck and unable to find the stationary points $x,y$ and hence the shortest distance.

Can someone help me to complete the solution of the problem ?

$\endgroup$
  • $\begingroup$ Now that you have a relation between $x$ and $y$, substitute in to the original equation and solve for one - you should end up with a quadratic. $\endgroup$ – platty Aug 10 '17 at 16:06
  • $\begingroup$ Bingo.....Got it. $\endgroup$ – learner Aug 10 '17 at 16:26
0
$\begingroup$

Let $M=\begin{pmatrix}1 & 4 \\ 4 & 7\end{pmatrix}$. We are looking for $$ \min_{(x\,y) M (x\, y)^T=225} x^2+y^2 $$ and $M$ is a symmetric matrix with eigenvalues $9,-1$, associated with the eigenvectors $(1,2)^T$ and $(-2,1)^T$. By the spectral theorem, the above minimum equals $$ \min_{9x^2-y^2=225} x^2+y^2 = \min_{9x^2-y^2=225} \frac{9x^2-y^2}{9}+\frac{10y^2}{9}=\color{blue}{25} $$ hence the wanted minimum distance equals $\color{blue}{5}$, attained by $\pm\left(\sqrt{5},2\sqrt{5}\right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.