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I have an question that is confusing me, I tried to search on internet but no clear answer... Questions : How can i determine the equation of a conic given 4 or 5 points example : Given the points A(1,1) B(0,2) C(-3,0) D(2,1) E(1,2) How to write the equation of conic passing through the points ( A,B,C,D ) ?

Tarek Chafei
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  • So you do not want to include point $E$ in your conic? – Mohammad Riazi-Kermani Jul 19 '19 at 15:15
  • Actually there is a related question to determine the conic that is passing through the 5 points...but i think knowing the method to obtain a conic from 4 points is the same as 5 points maybe. – Tarek Chafei Jul 19 '19 at 15:18
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    The vanishing of the 6x6 determinant $\begin{vmatrix}1&1&1&1&1&1\0&0&4&0&2&1\9&0&0&-3&0&1\4&2&1&2&1&1\1&2&4&1&2&1\x^2&xy&y^2&x&y&1\end{vmatrix}=2(x^2+2xy-14y^2-5x+40y-24)=0$ is an equation for the conic (a hyperbola) through the five points. – Jan-Magnus Økland Jul 19 '19 at 17:38
  • https://math.stackexchange.com/questions/2656488/equation-of-conic-given-five-points-on-it – Intelligenti pauca Jul 19 '19 at 18:29
  • @Jan-MagnusØkland solving 6x6 determinate needs long time...while i have to solve like this question in the exam... – Tarek Chafei Jul 19 '19 at 18:53
  • why your points do not have a third coordinate? – Vasili Jul 19 '19 at 20:10
  • @Vasya i’m talking about a conic like ( ellipse hyperbola ....)i’m not talking about things like ( ellipsoid and hyperboloid ...) so it must has two coordinates. – Tarek Chafei Jul 19 '19 at 20:16
  • If you don’t feel like computing this determinant, then solve the associated system of linear equations. If you’re lucky, that might be less work. There is no royal road to enlightenment. There are other methods that involve computing many more determinants, although they are of much smaller matrices. Frankly, if you’re in a time crunch, solving the linear system is probably the fasted and, most importantly, least error-prone method. – amd Jul 20 '19 at 00:01
  • Four points are not enough to determine a unique conic. You will either need other information about the conic or content yourself with a one-parameter family of them. – amd Jul 20 '19 at 00:02
  • Solving a linear system of five equations in six unknowns requires you make a choice about which unknown to leave free; it must be non-zero, and you can rarely tell which ones will be. The determinant method, however, doesn't care: it's a completely mechanical, decision-free process ... but, yes, it's time-consuming. Neither approach is really appropriate for a timed exam, unless the determinant has many zeros, and/or the system is next-to-trivial. (On balance, the system is probably faster, if you don't choose unwisely.) Such lengthy calculations are why we have computer algebra systems. :) – Blue Jul 20 '19 at 00:58

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The general equation of a plane conic has the form $\ ax^2 + bxy + cy^2 + dx + ey + f = 0\ $. Substituting the coordinates of any point lying on the conic into this equation gives you a homogeneous linear equation in the coefficients $\ a, b, c, d, e\ $ and $\ f\ $. Since there are six unknown coefficients you need at least five equations to determine them as unique multiples of a single non-zero parameter (and therefore to determine the conic uniquely). For the given example, we have: $$ \begin{matrix} \mathbf{A} \mbox{ on the conic }&\implies& a + b + c + d + e + f &=& 0\\ \mathbf{B} \ \mbox{ on the conic }&\implies& 4c + 2e + f &=& 0\\ \mathbf{C} \ \mbox{ on the conic }&\implies& 9a -3d +f &=& 0\\ \mathbf{D} \ \mbox{ on the conic }&\implies& 4a+ 2b + c + 2d + e + f &=& 0\\ \mathbf{E} \ \mbox{ on the conic }&\implies& a + 2b + 4c + d + 2e + f &=& 0\ . \end{matrix} $$ These equations have solution $$ \left(a,b,c,d,e,f\right) = -\frac{f}{24}\,\left(1, 2, -14, -5, 40, -24\right)\ , $$ So the conic passing through these five points has equation $$ x^2 +2xy-14y^2-5x+40y-24=0\ . $$

lonza leggiera
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