Find the equation of lines joining the origin and points of intersection $$x^2 +y^2+2xy =4$$ and $$3x^2 +5y^2 -xy =7$$
I tried solving the 1st equation which is actually a set of 2 parallel lines $x+y=\pm2$. And the second equation is an eclipse according to wolframalpha graph calculator. But I have only studied eclipse whose axes are parallel to the coordinate axes so I don't understand how to solve it. Putting $x=y-2$ doesn't really help as far as I could do.
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Let $f(x,y)=x^2+y^2+2xy-4$ and $g(x,y)=3x^2+5y^2-xy-7$, so that the two curves are given by the zero sets $f(x,y)=0$ and $g(x,y)=0$. Their intersections are also contained in the zero set of every linear combination of these function, and per “Plücker’s mu,” one that passes through the origin is $$f(0,0)g(x,y)-g(0,0)f(x,y) = 5x^2-18xy+13y^2 = (x-y)(5x-13y) = 0.$$ The two lines are thus $x=y$ and $5x=13y$.
The first equation gives $x+y=2$ or $x+y=-2$.
If $y=2-x$ then after substitute in the second we obtain: $$3x^2+5(2-x)^2-x(2-x)=7$$ or $$3x^2+5x^2-20x+20-2x+x^2-7=0$$ or $$9x^2-22x+13=0,$$ which gives two points: $(1,1)$ and $\left(\frac{13}{9},\frac{5}{9}\right)$.
In the case $y=-2-x$ we obtain $9x^2+22x+13=0$.
Finally, we'll obtain two equations: $y=x$ and $y=\frac{5}{13}x$.