This is from practice questions on quadratic equations in my book. I solved it in the following way:
Divide the equations by $x^2$
$\Longrightarrow c\left(\frac{y}{x}\right)^{2}+b\left(\frac{y}{x}\right)+a\ =\frac{d}{x} ...(i)$
$\Longrightarrow a\left(\frac{y}{x}\right)^{2}+c\left(\frac{y}{x}\right)+b\ =\frac{d}{x}...(ii)$
Subtract (ii) from (i) and by solving the resulting quadratic(by replacing $\frac{y}{x}$ with $t$) I get:
$$\frac{y}{x}=\frac{\left(c-b\right)\pm\sqrt{\left(b-c\right)^{2}-4\left(a-b\right)\left(c-a\right)}}{2(c-a)}$$
At this point I looked at the answer provided which was:
$x^2=y^2=d/a+b+c$ ;
$x/(c-a)=y/(a-b)=K$ where $K^2a(a^{2}+b^{2}+c^{2}-ab-bc-ca)=d$
My question is where does this solution come from and if my approach is correct how to go further?
Hint
You properly wrote$$t=\frac{y}{x}=\frac{\left(c-b\right)\pm\sqrt{\left(b-c\right)^{2}-4\left(a-b\right)\left(c-a\right)}}{2(c-a)}$$ but you did not notice that $$\left(b-c\right)^{2}-4\left(a-b\right)\left(c-a\right)=(b+c-2 a)^2$$ which makes $t=1$ and $t=\frac{b-a}{a-c}$.
This should make the problem much more simple now.