Identification of a conic section

Question

Consider the equation $(E)\hskip 5mm x^2+xy+ky^2+6x+10=0$. I am looking for conditions on $k$ for the graph of $(E)$ to be a circle or an ellipse.

Clearly, if it is a circle or an ellipse, its discriminant $\Delta=1-4k$ has to be negative,so $k>\frac{1}{4}$. But the converse is not necessarily true: when $k\leqslant \frac{1}{4}$, it is possible that the conic is degenerate.

It is actually the case: we write $q(x,y)=x^2+xy+ky^2+6x+10=(x+\frac{t}{2}+3)^2+P(y)$, where $P(y)=(k-\frac{1}{4})y^2-3y+1$. The discriminant of $P$ is $D=10-4k$, so if $k>\frac{5}{2}$, $P(y)>0$, so the graph of $(E)$ is empty. And it is a point when $k=\frac{5}{2}$. In the case $k<\frac{5}{2}$, $(E)$ has still a non empty graph, but may still be degenerate.

Drawing examples on computers suggest that the right condition on $k$ is $\frac{1}{4}<k<\frac{5}{2}$. But I am stuck on how to finish to prove it. Any hint?

Please note that this is something I have to explain to some students, that don't know anything on conic sections, past their equation in standard form and the formula to rotating conic sections (I tried to use rotations but the computations look very complicated because of the parameter $k$).

Answer 1

There is some information on a more complete classification of conic sections on this page.

With these notations, your equation: $$(E) \quad x^2+xy+ky^2+6x+10=0$$ has the associated matrix $A_E$ given by: $$A_E = \begin{pmatrix}1 &0.5&3\\0.5&k&0\\3&0&10\end{pmatrix}$$ Following the classification rules described in the article, you get the following conditions on the parameter $k$.

• Not degenerate: $$\det A_E \ne 0 \implies k \ne \frac{5}{2}$$
• Ellipse (incl. circle): $$\det A_{33} < 0 \implies k > \frac{1}{4}$$
• Real solutions (i.e. no imaginary ellipse): $$(A+C)\det A_E < 0 \implies -1 < k < \frac{5}{2}$$ So together that gives, indeed: $\frac{1}{4}< k < \frac{5}{2}$.