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So I have a question that says...

Find numbers $a,b,c,d,e$ and $f$ so that the five points $( 0,2 ),( -3 ,0 ),( 1, 5 ),( 1,1 ),( -1 ,1 )$ all lie on the conic $ax^2 + bxy + cy^2+dx+ey+f=0$ . Show, moreover, that $a,b,c,d,e$ and $f$ are uniquely determined up to a common factor.

I'm really confused as to how I would start this problem. I feel like I should be using matrix's at some point, but I need some equations with the points to start? Could some show/explain how I would get started going about this problem?

user6259845
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  • If you plug in the point $(0,2)$, you get the equation $(0)^2a+(0)(2)b+(2)^2c+(0)d+(2)e+f=0$ – JMoravitz Oct 27 '16 at 04:50
  • Just substitute the $(x,y)$ pairs into the equation and arbitrarily set $f=1$. You have five equations in five unknowns. – Parcly Taxel Oct 27 '16 at 04:51
  • @ParclyTaxel: It's safer not to set $f = 1$, since it's possible that the relevant conic might actually require $f = 0$. Instead, you should solve five equations in $6$ unknowns; this will solve the problem in the form indicated, i.e. the solutions set $(a,\dots,f)$ will be determined up to a scalar. – tracing Oct 27 '16 at 04:55
  • I reviewed - f doesn't have to be 0. But he brings up a good point. – Mark Oct 27 '16 at 05:20

3 Answers3

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Let $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$

we can solve the this system of linear equation:

$F(0,2)=F(-3,0)=F(1,5)=F(1,1)=F(-1,1)=0$ and get $$ \begin{bmatrix} a \\ b \\ c \\ d \\ e \\ f \\ \end{bmatrix} =\begin{bmatrix} (-7/6)f \\ (19/6)f \\ (-2/3)f \\ (-19/6)f \\ (5/6)f \\ f \\ \end{bmatrix}$$ where $f$ can not be $0$

Acuzio Leigh
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Plugging each point into the given equation gives \begin{align*} 4 \, c + 2 \, e + f &= 0 \\ 9 \, a - 3 \, d + f &= 0 \\ a + 5 \, b + 25 \, c + d + 5 \, e + f &= 0 \\ a + b + c + d + e + f &= 0 \\ a - b + c - d + e + f &= 0 \end{align*} This system can be written in matrix form as $A\vec x=\vec 0$ where \begin{align*} A &= \left[\begin{array}{rrrrrr} 0 & 0 & 4 & 0 & 2 & 1 \\ 9 & 0 & 0 & -3 & 0 & 1 \\ 1 & 5 & 25 & 1 & 5 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 & 1 & 1 \end{array}\right] & \vec x &= \left[\begin{array}{r} a \\ b \\ c \\ d \\ e \\ f \end{array}\right] \end{align*} Now, row reduce $A$ to obtain $$ \DeclareMathOperator{rref}{rref}\rref(A)= \left[\begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & \frac{7}{6} \\ 0 & 1 & 0 & 0 & 0 & -\frac{19}{6} \\ 0 & 0 & 1 & 0 & 0 & \frac{2}{3} \\ 0 & 0 & 0 & 1 & 0 & \frac{19}{6} \\ 0 & 0 & 0 & 0 & 1 & -\frac{5}{6} \end{array}\right] $$ This implies that our original system is equivalent to \begin{align*} a + \frac{7}{6} \, f &= 0 & b - \frac{19}{6} \, f&= 0 & c + \frac{2}{3} \, f&= 0 & d + \frac{19}{6} \, f&= 0 & e - \frac{5}{6} \, f&= 0 \end{align*}

Brian Fitzpatrick
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Solving the linear system, as others have shown, is perhaps the best way to go if you're doing this by hand and/or for the first time (and if you want to convince yourself of uniqueness-up-to-common-factor), but I'll just mention this ...

An equation of a conic through five points ---say, $A$, $B$, $C$, $D$, $E$--- is given by this determinant relation (where $P_x$ and $P_y$ are, respectively, the $x$- and $y$-coordinates of $P$):

$$\left|\begin{matrix} x^2 & y^2 & x y & x & y & 1 \\ A_x^2 & A_y^2 & A_x A_y & A_x & A_y & 1 \\ B_x^2 & B_y^2 & B_x B_y & B_x & B_y & 1 \\ C_x^2 & C_y^2 & C_x C_y & C_x & C_y & 1 \\ D_x^2 & D_y^2 & D_x D_y & D_x & D_y & 1 \\ E_x^2 & E_y^2 & E_x E_y & E_x & E_y & 1 \\ \end{matrix}\right| = 0 \tag{$\star$}$$

So, in theory, one "simply" expands the determinant and reads-off the various coefficients. Of course, no one wants to expand a $6\times 6$ determinant by hand, but $(\star)$ comes in (ahem) handy if you have a computer algebra system available to do the symbol crunching.

Blue
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