How to prove that any five points, of which no 3 are colinear, that there is a single conic that passes through all of them?
(I have to start out with the equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ but I don't know what to do afterward.)
[Math] Why 5 points determine a conic
analytic geometryconic sectionsgeometry
Related Solutions
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.
Use Brianchon's theorem as in the following diagram:
If $F$ is the point of contact of the tangent $CD$, then by Brianchon's theorem the three diagonals of the hexagon $ABCFDE$ are concurrent at a point $O$.
So we can intersect $BD$ and $CE$ to find $O$, and the contact point $F$ lies on the line $AO$. Similarly we can find the four other contact points.
This gives us five points, from which we can construct the conic.
Best Answer
Here's the outline of one method that uses some linear algebra. $\def\Ker{\mathop{\mathrm{Ker}}}$
Consider the set of polynomials \begin{align} V = \{ Ax^2+Bxy+Cy^2+Dx+Ey+F \mid A,\ldots,F\in\mathbf R \} \end{align} and show that this is a $6$-dimensional vector space.
Show that for two vectors $f(x,y)$ and $g(x,y)$ in $V\setminus\{0\}$, the conics defined by $f(x,y)=0$ and $g(x,y)=0$ are the same if, and only if, $f$ and $g$ are colinear.
For any point $P=(a,b)$ in the plane, consider the map $\varphi_P:V\to\mathbf R$ defined by $\varphi_P(f) = f(a,b)$ for any $f(x,y)\in V$. Show that it is a surjective linear map. From the surjectivity, deduce that \begin{align} \dim(\Ker\varphi_P) = 5. \end{align}
Show that a point $P$ is on the conic $f(x,y)=0$ if, and only if, $\varphi_P(f) = 0$.
Consider five points $P_1,\ldots,P_5$. Show that \begin{align} \dim\left(\bigcap_{i=1}^5 \Ker\varphi_{P_i}\right) = 1\end{align} if, and only if, no three of the points are colinear.
Conclude.