1. Given parabola and line

The parabola is $x^2=4y$ which can be written as $y=\frac{x^2}{4}.$

The chord is cut by the line $x-\sqrt{2}\,y+4\sqrt{2}=0.$

We need the length of the chord, i.e. the distance between the two intersection points of the line and the parabola.

2. Find the points of intersection

From the line, $x-\sqrt{2}y+4\sqrt{2}=0$ $\sqrt{2}y=x+4\sqrt{2}$ $y=\frac{x}{\sqrt{2}}+4.$

Now substitute into the parabola equation: $\frac{x^2}{4}=\frac{x}{\sqrt{2}}+4.$

Multiply by $4$: $x^2=2\sqrt{2}x+16.$

So, $x^2-2\sqrt{2}x-16=0.$

Solve this quadratic: $x=\frac{2\sqrt{2}\pm\sqrt{(2\sqrt{2})^2+64}}{2}$ $=\frac{2\sqrt{2}\pm\sqrt{8+64}}{2}$ $=\frac{2\sqrt{2}\pm\sqrt{72}}{2}$ $=\frac{2\sqrt{2}\pm 6\sqrt{2}}{2}.$

Hence, $x=4\sqrt{2}\quad \text{or} \quad x=-2\sqrt{2}.$

Now find corresponding $y$ values using $y=\frac{x^2}{4}:$

• For $x=4\sqrt{2}$, $y=\frac{(4\sqrt{2})^2}{4}=\frac{32}{4}=8.$ So one point is $P(4\sqrt{2},8).$

• For $x=-2\sqrt{2}$, $y=\frac{(-2\sqrt{2})^2}{4}=\frac{8}{4}=2.$ So the other point is $Q(-2\sqrt{2},2).$

3. Find the length of the chord

Using the distance formula between $P(4\sqrt{2},8)$ and $Q(-2\sqrt{2},2)$:

$PQ=\sqrt{(4\sqrt{2}-(-2\sqrt{2}))^2+(8-2)^2}$ $=\sqrt{(6\sqrt{2})^2+6^2}$ $=\sqrt{72+36}$ $=\sqrt{108}$ $=6\sqrt{3}.$

4. Compare with options

The chord length is $\boxed{6\sqrt{3}}.$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B
Derived answer: B

They agree.