Problem:
Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.
My attempt:
Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.
$$b=-a \\c=4a \\d=e=0 \\f=-4a$$
Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 \ \ (1)$$
My question:
Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.
If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?
Best Answer
Any second-degree curve equation can be written as $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}$$ or $$ax^2+2hxy+by^2+2gx+2fy+c=0\tag{2}$$ where $$A,B,C,D,E,E,a,b,c,f,g,h\in\mathbb R$$
To find type of conic and nature of conic we use $\Delta$ and is given by $$\Delta=\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}$$ $$=abc+2fgh-af^2-bg^2-ch^2\tag{3}$$
If $\Delta$ is $0$, it represents a degenerate conic section. Otherwise, it represents a non-degenerate conic section.
Also, the type of conic section that the above equation represents can be found using the discriminant of the equation, which is given by $B^2-4AC$.
Conditions regarding the quadratic discriminant are as follows:
If $\Delta=0$:
$\bullet$ If $h^2-ab\gt0$, the equation represents two distinct real lines.
$\bullet$ If $h^2-ab=0$, the equation represents parallel lines.
$\bullet$ If $h^2-ab\lt0$, the equation represents non-real lines.
If $\Delta\neq0$:
$\bullet$ If $B^2-4AC\gt0$, it represents a hyperbola and a rectangular hyperbola $(A+C=0)$.
$\bullet$ If $B^2-4AC=0$, the equation represents a parabola.
$\bullet$ If $B^2-4AC\lt0$, the equation represents a circle $(A=C,B=0$) or an ellipse $(A\neq C)$. For a real ellipse, $\Big(\frac\Delta{a+b}\lt0\Big)$.
So for the given case, the equation of conic is (after putting all the points in the general equation of conic and finding all the coefficient) $$ax^2-axy+4ay^2-4a=0$$
Comparing above equation with equation (1) and (2) we get the values of required coefficient as $A=a$, $B=-a$ and $C=4a$ so we get the value of $B^2-4AC=-3a^2$ and also from equation (3) we get $\Delta=-15a^2$. Since $\Delta\neq0$ and $B^2-4AC\lt0$ and also $A\neq C$, so as we know, this is the condition of an ellipse.
$\therefore$ Conic $ax^2-axy+4ay^2-4a=0$ is a real ellipse.