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Given two second degree conic equations: $ax^2+by^2+cx+dy+e=0$ and $fx^2+gy^2+hx+iy+j=0$ [All coefficients are real]

To solve these equation for $x$ and $y$ a direct substitution yields a polynomial of fourth degree in $x$ (or $y$) as: $kx^4+lx^3+mx^2+nx+o=0$ Is there any easier way to evaluate the coefficients of fourth degree Polynomial $(k,l,m,n,o)$ in $x$ (or $y$) in terms of $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$, $i$, $j$.

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    easier than a direct substitution of one variable into the second equation2012-03-05

2 Answers 2

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Consider your two polynomials as polynomials $A(y),B(y)$ with coefficients that are polynomials in $x$, and compute the resultant of the two polynomials $A(y),B(y)$. That will give you the polynomial in $x$ that you want. You can calculate the resultant using the Sylvester determinant. Information about these two concepts is widely available on the web and elsewhere.

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$g(ax^2+by^2+cx+dy+e)=0$

$-b(fx^2+gy^2+hx+iy+j)=0$

Sum two equations and collect y in one side. ($y^2$ term will be zero )

$y= \frac{(bf-ga)x^2+(bh-gc)x+(bj-ge)}{gd-bi}=\frac{(bf-ga)}{gd-bi}x^2+\frac{(bh-gc)}{gd-bi}x+\frac{(bj-ge)}{gd-bi}$ and then put in any of equation that you gave and you will get the coefficients of fourth degree Polynomial.