Question
The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.
Attempt
Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$.
Then how it is ellipse?
Any hints or suggestion?
On
The matrix $$ \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} $$ defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2\pm\sqrt 5$. The eigenvector associated to the positive eigenvalue is $(\sqrt5-1,2)$. Therefore the vertices are obtained by solving $$ \left\{\begin{array}{l} x^2+4xy+3y^2=4 \\ (x,y) = t(\sqrt5-1,2) \\ \end{array}\right. $$ The solutions are $$ x_0 = \sqrt{\frac{14}{\sqrt{5 }}-6} \qquad y_0 = \sqrt{\frac{6}{\sqrt{ 5}}-2} $$ and the opposite point $-(x_0,y_0)$.
Therefore the distance from the center (which is the origin) is $$ |(x_0,y_0)| = \sqrt{x_0^2+y_0^2} = 2 \sqrt{\sqrt5-2} . $$
The quadratic form is $$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$ The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic
Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $\tan A = 1$ and $ \tan B = 3.$ The trig formula for the bisecting angle is just $$ \tan \left( \frac{A+B}{2} \right) = \frac{\sin A + \sin B}{\cos A + \cos B} $$