# Using Lagrange's method find the shortest distance from the origin to the hyperbola

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 ?

• 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. – platty Aug 10 '17 at 16:06
• Bingo.....Got it. – learner Aug 10 '17 at 16:26

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 $\pm\left(\sqrt{5},2\sqrt{5}\right)$.