Given $E: 2x^2-xy+y^2+y=4$ and $H: 2x^2-y^2=1$, find intersection.

All the times I had to deal with problems of intersection I would always use substitution method, but now it seems a tedious way of dealing with this problem. Is there a better approach?

  • $\begingroup$ Do you have any tool so that you are able to solve a $4$ degree equation. $\endgroup$ Jun 22, 2017 at 6:17
  • $\begingroup$ Yes, at least I hope so. $\endgroup$
    – maq
    Jun 22, 2017 at 6:19

2 Answers 2


I'd follow this answer of mine.

  1. Turn your conics into symmetric matrices: $$A=\begin{bmatrix}4&-1&0\\-1&2&1\\0&1&-8\end{bmatrix}\qquad B=\begin{bmatrix}2&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$$ Note that I scaled your first equation by a factor of $2$ because symmetrically distributing the off-diagonal terms divides them by two and I wanted to avoid fractions.

  2. Find linear combinations with zero determinant. $$\det(\lambda A+\mu B)=-60\lambda^3-9\lambda^2\mu+16\lambda\mu^2+2\mu^3$$ This does not factor so you can't expect rational solutions. In particular zero is not a solution for either $\lambda$ or $\mu$ (although $\lambda=\mu=0$ is a solution but we ignore that). So set $\lambda=1$ and compute $\mu$ for that: $$\mu_1\approx-8.10\qquad\mu_2\approx-1.88\qquad\mu_3\approx1.97$$ The exact solutions are tedius to write out, so I'll stick to approximate numbers.

  3. $A+\mu_iB$ is a degenerate conic formed by a pair of lines passing through the points of intersection. I'll continue with $\mu_1$ for my numeric examples. $$C_1=A+\mu_1B\approx\begin{bmatrix}-12.2&-1&0\\-1&10.1&1\\0&1&0.0982\end{bmatrix}$$

  4. The adjugate of that is $$\operatorname{adj}C_1\approx\begin{bmatrix}-0.00805&0.0982&-1\\0.0982&-1.20&12.2\\-1&12.2&-124\end{bmatrix}$$ It has rank 1, so all rows and all columns are multiple of one another. Take any non-zero row or column of that, e.g. the first row, and you have the homogeneous coordinates where the two lines intersect. (Divide by the last coordinate if you want to have regular $(x,y)$ coordinates but you don't need that.) Now use these coordinates to form an anti-symmetric matrix $$P_1\approx\begin{bmatrix}0&-1&-0.0982\\1&0&-0.00805\\0.0982&0.00805&0\end{bmatrix}$$ Then consider $$C_1+\lambda P_1\approx\begin{bmatrix}-12.2&-1-\lambda&-0.0982\lambda\\-1+\lambda&10.1&1-0.00805\lambda\\0.0982\lambda&1+0.00805\lambda&0.0982\end{bmatrix}$$ Take any $2\times2$ minor of this, e.g. $$\begin{vmatrix}-12.2&-1-\lambda\\-1+\lambda&10.1\end{vmatrix}\approx\lambda^2-124\overset!=0$$ From this you can conclude that $\lambda\approx\pm11.1$. Choosing the positive solution (an arbitrary choice) you get $$C_1+\lambda P_1\approx\begin{bmatrix}-12.2&-12.1&-1.09\\10.1&10.1&0.910\\1.09&1.09&0.0982\end{bmatrix}$$ This matrix has again rank one, so its rows are multiples of one another, and its columns are multiples of one another. Pick any non-zero row and any non-zero column, and you have the equations of two of the lines. $$g_1\approx[-12.2:-12.1:-1.09]\qquad h_1\approx[-12.2:10.1:1.09]$$.

  5. Repeat these steps for $\mu_2$. I wont print this here, but in the end you obtain $$g_2\approx[0.248:-1.20:1.23]\qquad h_2\approx[0.248:-0.799:-1.23]$$ Now intersect one of the lines from roun $1$ with one from round $2$ by computing the corss product to obtain one of your points of intersection. Divide by the last coordinate to dehomogenize. \begin{align*}q_1&=g_1\times g_2\approx[-16.3:14.7:17.7]\to(-0.921, 0.835)\\q_2&=g_1\times h_2\approx[14.1:-15.3:12.8]\to(1.10, -1.20)\\q_3&=h_1\times g_2\approx[13.8:15.3:12.1]\to(1.14, 1.26)\\q_4&=h_1\times h_2\approx[-11.6, -14.7, 7.23]\to(-1.61, -2.04)\end{align*}

You can use WolframAlpha to check this result.

In case you want exact results instead of those numeric approximations, note that all the solutions satisfy

\begin{align*} 14x^4 + 4x^3 - 41x^2 - 2x + 26 &=0 \\ 7y^4 + 8y^3 - 23y^2 - 12y + 18 &=0 \end{align*}

as you might find by using elimination techniques such as resultants (not shown here). But these polynomials don't factor, so a formulation using radicals would be fairly complicated. And pairing the appropriate solutions would still be left as a follow-up task if you didn't have the derivation above, so starting with those equations isn't always useful.

  • $\begingroup$ The Lex grobner basis from M2: matrix {{7*y^4+8*y^3-23*y^2-12*y+18, 6*x-7*y^3-8*y^2+11*y+6}} $\endgroup$ Jan 27, 2023 at 14:58

You can rewrite the first equation as $2y^2-xy+y=3$ (using $2x^2-y^2=1$).
Rearranging it to $2y^2-3=y (x-1)$ and squaring it, we have $(2y^2-3)^2=y^2 (x-1)^2$.
Plugging $y^2=2x^2-1$, we get the simplified equation in $x$ as $14x^4+4x^3-41x^2-2x+26=0$ (Hoping all my algebra is correct).

The roots in approximate form are $-1.61,-0.92,1.10,1.14$. This has got $4$ roots. But as we have squared, we also have extraneous roots. Thus after getting values of $y$ from corresponding $x$ using the equation of hyperbola, we need to plug it again in the ellipse equation as the roots of $4$ degree may or may not be the points of intersection of ellipse and hyperbola.


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