Normal to the curve $y=x^2$ forming the shortest chord 
Find the normal to the curve $y=x^2$ forming the shortest chord

Normal at point $P(x,y)=(t,t^2)$ is $x+2ty=t+2t^3$, which meets the point $Q(m,m^2)$
$$
m+2tm^2=t+2t^3\implies m-t=-2t(m^2-t^2)=-2t(m-t)(m+t)\\
t+m=\frac{-1}{2t}\implies m=-t-\frac{1}{2t}\\
T=\Delta^2=(x_1-x_2)^2+(y_1-y_2)^2\\
=(t-m)^2+(t^2-m^2)^2=(t-m)^2[1+(t+m)^2]=(2t+\frac{1}{2t})^2\big[1+\frac{1}{4t^2}\big]\\
=(\frac{4t^2+1}{2t})^2(\frac{4t^2+1}{4t^2})=\bigg[\frac{4t^2+1}{2t}\bigg]^3\\
\frac{dT}{dt}=3\bigg[\frac{4t^2+1}{2t}\bigg]^2.\frac{2t(8t)-(4t^2+1).2}{4t^2}=3\bigg[\frac{4t^2+1}{2t}\bigg]^2.\frac{16t^2-8t^2-2}{4t^2}\\
=3\bigg[\frac{4t^2+1}{2t}\bigg]^2.2.\frac{4t^2-1}{(2t)^2}=0\\
\implies t^2=\frac{1}{4}\implies t=\pm\frac{1}{2}\\
N:x+y=3/4
$$
But my reference gives the solution $y=\pm\dfrac{x}{\sqrt{2}}+1$, which is asked Which normal to $=^2$ forms the shortest chord?, but I really can't figure out what is going wrong with my attempt ?

$\left(\frac{4t^2+1}{2t}\right)^2\frac{4t^2+1}{4t^2}=\frac{(4t^2+1)^3}{(2t)^4}$ and not $\left(\frac{4t^2+1}{2t}\right)^3$. With this correction it should continue like user207119's answer to the linked question.
