$$ F(X,Y)=(t_1^2+t_2^2+2t_1t_2X)^2Y^2-2t_3^2(t_1^2X+t_2^2X+2t_1t_2)Y+t_3^4 $$
I want to know the point $(X,Y)$ which satisfied with $F(X,Y)=0$. Now, $t_1,t_2,t_3$ are positive numbers.
By numerical calculation, I noticed $F(1,(\frac{t_3}{t_1+t_2})^2)=0$ when $\frac{t_3}{t_1+t_2}<1$.
I want to know how to know this result analytically. Is it possible??
Hint: consider $F$ as a quadratic in $Y$ and note that its reduced discriminant is:
$$ \frac{1}{4}\Delta = t_3^4 (t_1^2 X + t_2^2 X + 2 t_1 t_2)^2 - t_3^4(t_1^2+t_2^2+2t_1 t_2 X)^2=t_3^4 (t_1^2 - t_2^2)^2 (X^2 - 1) $$
Therefore $\,F(X,Y)=(t_1^2+t_2^2+2t_1t_2X)^2(Y-Y_1)(Y-Y_2)\,$ where:
$$ Y_{1,2} = \frac{t_3^2(t_1^2X+t_2^2X+2t_1t_2) \pm t_3^2(t_1^2-t_2^2) \sqrt{X^2-1} \big/ 2}{(t_1^2+t_2^2+2t_1t_2X)^2} $$
[ EDIT ] For $\,X=1\,$ the above reduces to $\,Y_1=Y_2=t_3^2 / (t_1+t_2)^2\,$ so:
$$ F(1,Y)=(t_1+t_2)^4\big(Y-t_3^2 / (t_1+t_2)^2\big)^2 = \big((t_1+t_2)^2 Y - t_3^2\big)^2 $$