<s>[INST] Find the equation of the chord joining points $p$ and $q$ on the parabola $x=2t$, $y=t^2$ if $p$ and $q$ are the roots of the equation $t^2-4t+2=0$ I have the answer but do not know the process in achieving it.
Find the equation of the chord joining points $p$ and $q$ on the parabola $x=2t$, $y=t^2$ if $p$ and $q$ are the roots of the equation $t^2-4t+2=0$.
 [/INST] We have $$x=2t,  \ y=t^2$$ $$\implies x^2=4y$$ 
Now, solving $t^2-4t+2=0$, we get $$t=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$$ $$t=\frac{4\pm 2\sqrt 2}{2}=2\pm\sqrt 2$$
Setting the values of $t$, we get the coordinates of the points P & Q as follows $$P(2t, t^2)\equiv (4+ 2\sqrt 2,\ 6+4\sqrt 2 )$$ & 
$$Q(2t, t^2)\equiv (4- 2\sqrt 2,\ 6-4\sqrt 2 )$$
I hope you can find out the equation of the chord PQ passing through the points $P(4+ 2\sqrt 2,\ 6+4\sqrt 2 )$ & $Q(4- 2\sqrt 2,\ 6-4\sqrt 2 )$ using formula
$$y-(6+4\sqrt 2)=\frac{(6+4\sqrt 2)-(6-4\sqrt 2)}{(4+2\sqrt 2)-(4-2\sqrt 2)}(x-(4+ 2\sqrt 2))$$
$$y-(6+4\sqrt 2)=2(x-4-2\sqrt 2)$$
$$y=2x-2$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the chord PQ:}\ \ \color{blue}{2x-y-2=0}}}$$
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