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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.

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Using your idea, we get $$\left(mx+\frac 12\right)^2=(1-t^2)x^2+\frac 14,$$ i.e. $$x(m^2x+m-x+t^2x)=0$$ Since $x\not=0$, we get $$x=\frac{m}{1-m^2-t^2},\quad y=mx+\frac 12=\frac{m^2+1-t^2}{2(1-m^2-t^2)}$$ where $m\in\mathbb Q$ has to satisfy $1-m^2-t^2\not=0,m\not=0$ and $m^2+1-t^2\not=0$.

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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 $$x=\frac{m}{1-m^2-t^2},\qquad y=\frac{m^2}{1-m^2-t^2}+\frac12=\frac{1+m^2-t^2}{2(1-m^2-t^2)}.$$

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