Solving $\frac{\alpha^2}{(t_1-x)^2}+\frac{\beta^2}{(t_2-x)^2}=1$

Question

For a given $\alpha$ and $\beta$ such that $\alpha^2+\beta^2=1$ and $t_1,t_2>0, t_1\neq t_2$, I'm trying to solve the following equation: $$\frac{\alpha^2}{(t_1-x)^2}+\frac{\beta^2}{(t_2-x)^2}=1$$ I tried to calculate delta for a forth degree equation but it's too complicated is there a better way to solve it?

Answer 1

$$\frac{\alpha^2}{(t_1-x)^2}+\frac{\beta^2}{(t_2-x)^2}=1 \Rightarrow \frac{\alpha^2}{(t_1-x)^2}+\frac{\beta^2}{(t_2-x)^2}= \alpha^2+\beta^2$$

So

$$(t_1-x)^2=1 \text{ and } (t_2-x)^2=1$$