I want to solve this quadratic system of equations in three variables $\alpha,\beta,\lambda\in \mathbb{R}$:
$$2a\alpha + 2b\beta + d + 2f\lambda\alpha + 2g\lambda\beta=0$$
$$2b\alpha + 2c\beta + e + 2g\lambda\alpha + 2h\lambda\beta=0$$
$$ f\alpha^2 + 2g\alpha\beta + h\beta^2 - \Delta^2 = 0$$
For any $a,b,c,d,e,f,g,h,\Delta\in \mathbb{R}$ with the condition that $a,c,f,h\geq 0$
My approach is to eliminate $\lambda$ from the first two equations, which leaves me with the following simultaneous quadratic equations.
$$2(ag-fb)\alpha^2 + 2(ah-cf)\alpha\beta + 2(bh-cg)\beta^2 + (dg-ef)\alpha + (dh-eg)\beta=0$$ $$ f\alpha^2 + 2g\alpha\beta + h\beta^2 - \Delta^2 = 0$$
What is the best approach to solve these equations?
I tried making a parametric substitution in $t$ into the second equation of $\alpha=\frac{z_1t}{1+t^2}$, $\beta=\frac{1 + z_2t + z_3t^3}{1+t^2}$ for some $z_i$ constant coefficients but when I solved for the coefficients $z_1=z_2=0$.
I would also be interested if you have a different approach for solving the initial problem in three variables.
What I should do is to use equations $(1)$ and $(2)$ which are linear in $\alpha$ and $\beta$ and solve them for $(\alpha,\beta)$ in terms of $\lambda$ and the various constants. The expressions are quite simple.
Now, plug their expressions in $(3)$, reduce to same denominator. You will have a quartic equation in $\lambda$.