Given objects

• Line: $x-y=1$
• Parabola: $x^2=2y \Rightarrow y=\dfrac{x^2}{2}$

Write a general point on the parabola Let a point on $x^2=2y$ be $P\left(t,\frac{t^2}{2}\right).$

Distance from a point to the line For line $x-y-1=0$, distance of point $(x_1,y_1)$ is $d=\frac{|x_1-y_1-1|}{\sqrt{1^2+(-1)^2}}=\frac{|x_1-y_1-1|}{\sqrt{2}}.$

So for $P\left(t,\frac{t^2}{2}\right)$, $d(t)=\frac{\left|t-\frac{t^2}{2}-1\right|}{\sqrt{2}}.$

Simplify the expression inside modulus $t-\frac{t^2}{2}-1= -\left(\frac{t^2}{2}-t+1\right).$ Hence $\left|t-\frac{t^2}{2}-1\right|=\left|\frac{t^2}{2}-t+1\right|.$

Now, $\frac{t^2}{2}-t+1=\frac{1}{2}(t^2-2t+2)=\frac{1}{2}\big((t-1)^2+1\big).$

Since $((t-1)^2+1)>0$ for all real $t$, modulus is unnecessary: $\left|\frac{t^2}{2}-t+1\right|=\frac{1}{2}\big((t-1)^2+1\big).$

Therefore,

Minimize the distance Since $(t-1)^2\ge 0$, the minimum occurs at $t=1.$

Then $d_{\min}=\frac{0+1}{2\sqrt{2}}=\frac{1}{2\sqrt{2}}.$

Check with options $\frac{1}{2\sqrt{2}}$ corresponds to Option B.

Alternative geometric check The line $x-y=1$ can be written as $y=x-1$. Intersections with parabola satisfy $x^2=2(x-1) \Rightarrow x^2-2x+2=0,$ whose discriminant is $\Delta =(-2)^2-4(1)(2)=4-8=-4<0.$ So they do not intersect, and the minimum distance must be positive, consistent with the result above.