Parabola stuff

$\large{Hints:}$

Equation of normal at $(t,t^{2})$ is
$\boxed{y+\frac{x}{2t}=t^{2}+\frac{1}{2} }$

Point of intersection of this normal and parabola $y=x^{2}$ is

$x^{2}+\frac{x}{2t}=t^{2}+\frac{1}{2}.$

One root of this equation is we know $x=t$ and sum of root is $-\frac{1}{2t}$ $$ So other root is $x=-t-\frac{1}{2t}$ So here y- coordinate is $t^{2}+\frac{1}{4t^{2}}+1.$

So distance between $(t,t^{2})$ and $(-t-\frac{1}{2t},t^{2}+\frac{1}{4t^{2}}+1)$ is to be shortest.

Distance is $\displaystyle\sqrt{(1+4t^{2})(1+\frac{1}{4t^{2}})^{2}}$

Differentiating inner stuff we can easily get $t=\pm(\frac{1}{\sqrt2}).$

So there are two equations

$y=\pm(\frac{1}{\sqrt2})x+1$ .

In each case $\boxed{m^{2}+c^{2}=1.5}$