Finding all the rational points of the curve $x^2 -2y^2 =1$ This question was left as an exercise in my class of number theory and I want to verify my solution.

Question: Let $C$ be a curve given by $x^2 -2 y^2=1$. Find all the rational points on $C$.

Attempt: $( 1,0 )$ is a solution and let m be the slope of the line passing through $(1,0)$ . Now the equation of the line is $y-0 = t(x-1)$ (say $L$). So, $x= 1+y/t$.
As, the curve is an hyperbola, so this line say L will intersect the curve at least $1$ more time.
Putting $x= y/t+1$ in $x^2 -2y^2 =1 $, I get y = $\frac {2t } { 2 t^2 -1} $ and $x=  + \frac{ 2t^2 +1} {2t^2-1} , - \frac{ 2t^2 +1} {2t^2-1}$. So, these are the rational points of the curve along with $(1,0)$.
Is my solution fine?
 A: What you did is basically OK. However, there are several places that could be improved.
A detailed solution.
If $x=1$, then $y=0$.
Suppose $(x,y)$ is a rational solution $(x,y)$ with $x\not=1$. Let the slope of the line through $(1,0)$ and $(x,y)$ be $t=\frac{y-0}{x-1}=\frac y{x-1}$, which must be a rational number.
$t=\frac y{x-1}$ implies $y=t(x-1)$. Putting $y=t(x-1)$ in $x^2-2y^2 =1$, we get $$(x-1)((2t^2-1)x-(2t^2+1))=0.$$ Since $x\not=1$, we get $x=\frac{ 2t^2 +1} {2t^2-1}$ and then y = $\frac {2t } { 2 t^2 -1} $.
Note that when $t$ is a rational number, $2t^2-1\not=0$ and both $\frac{2t^2+1}{2t^2-1}$ and $\frac{2t}{2t^2-1}$ are rational numbers.
So, all rational points are $(1,0)$ and $(\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$.
Wait, don't we also have $(-\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$?
Yes, $(-\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$ are also rational points on $C$. However, they do not provide any point that is not $(1,0)$ or $(\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for some rational $t$. This fact is clear since the meticulous solution process above must have found all rational points on $C$.
It turns out that there are many parametrized solutions.
Besides the one given by $(1,0)$ and $(\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$, all rational points on $C$ can also be $(-1,0)$ and $(-\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$, where $t$ can be understood as the negative of the slope of the line through $(-1,0)$ and $(x,y)$.
For anther example, consider $(3,2)$ on $C$. Let $t$ be the slope of the line through $(3,2)$ and $(x,y)$. We can find that all rational points on $C$ are $(3,\pm2)$ and $(\frac{6t^2-8t+3}{2t^2-1}, \frac{-4t^2+6t-2}{2t^2-1})$ for all rational $t$.
More generally, suppose $(a,b)$ is a rational point on $C$. Then all rational points on $C$ are $(a, \pm b)$ and $(\frac{a(2t^2+1)-4bt}{2t^2-1}, \frac{-b(2t^2+1)+2at}{2t^2-1})$ for all rational $t$.
