Rational solutions to a quadratic-type equation. For any fixed rational $t$, are there any nonzero rational solutions to the equation
$$y^{2} = (1-t^2)x^2 + \frac{1}{4} \text{ ?} $$
My attempt: consider the line $L$ whose gradient is $m$ and passes through the point $(0, \frac{1}{2})$. Then equation of $L$ is $y=mx + 1/2$, then consider its intersection with the quadratic in question and simplify. However, i'm yet to find any idea beyond this stage.

To simplify, rewrite the original equation as $(2y)^2+4(t^2-1)x^2=1$ and substitute $2y$ with $2mx+1$:
\begin{align}&(2mx+1)^2+4(t^2-1)=4(m^2+t^2-1)x^2+4mx+1=1\\
\iff & 4x\bigl((m^2+t^2-1)x+m\bigr)=0,
\end{align}
so the non-zero solution is
$$x=\frac{m}{1-m^2-t^2},\qquad y=\frac{m^2}{1-m^2-t^2}+\frac12=\frac{1+m^2-t^2}{2(1-m^2-t^2)}.$$
