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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 what? Which 4th degree polynomial are we talking about? Just the product of the two you have? – Gerry Myerson Mar 5 '12 at 4:04
easier than a direct substitution of one variable into the second equation – Abhinav Mar 5 '12 at 4:12

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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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.

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