An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points?

The two foci of the required ellipse are given as $$(0,2)$$ and $$(0,-2)$$. Because both lie on the $$y$$-axis and are symmetric about the origin, the centre of the ellipse is clearly $$O(0,0)$$ and the major axis is vertical.

For an ellipse whose centre is the origin and whose major axis is vertical, the standard form is

$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad\text{with }a\gt b\gt 0.$$

In this form the distance of each focus from the centre is denoted by $$c$$, and the coordinates of the foci are $$(0,\pm c)$$. The relationship among the three semi-lengths is the well-known formula

$$c^{2}=a^{2}-b^{2}.$$

Now, from the given foci $$(0,2)$$ and $$(0,-2)$$ we see that the distance of a focus from the centre is $$c=2$$, so

$$c^{2}=2^{2}=4.$$

The question states that the minor axis has length $$4$$. The minor axis equals $$2b$$, so

$$2b=4\;\Longrightarrow\;b=2\;\Longrightarrow\;b^{2}=2^{2}=4.$$

Substituting $$c^{2}=4$$ and $$b^{2}=4$$ into $$c^{2}=a^{2}-b^{2}$$ gives

$$4=a^{2}-4 \;\Longrightarrow\; a^{2}=4+4=8.$$

Thus the explicit equation of the ellipse is

$$\frac{x^{2}}{4}+\frac{y^{2}}{8}=1.$$

We must check which of the listed points satisfies this equation exactly.

Option A: $$(1,\,2\sqrt{2})$$

Substituting, we get $$\frac{1^{2}}{4}+\frac{(2\sqrt{2})^{2}}{8} =\frac14+\frac{8}{8} =\frac14+1 =\frac54\gt 1.$$ The left-hand side exceeds 1, so the point lies outside the ellipse.

Option B: $$(2,\,\sqrt{2})$$

Substituting, we get $$\frac{2^{2}}{4}+\frac{(\sqrt{2})^{2}}{8} =\frac{4}{4}+\frac{2}{8} =1+\frac14 =\frac54\gt 1.$$ Again the value is greater than 1, so this point is also outside the ellipse.

Option C: $$(\sqrt{2},\,2)$$

Substituting, we obtain $$\frac{(\sqrt{2})^{2}}{4}+\frac{2^{2}}{8} =\frac{2}{4}+\frac{4}{8} =\frac12+\frac12 =1.$$ The left-hand side equals 1 exactly, hence this point lies on the ellipse.

Option D: $$(2,\,2\sqrt{2})$$

Substituting, we get $$\frac{2^{2}}{4}+\frac{(2\sqrt{2})^{2}}{8} =\frac{4}{4}+\frac{8}{8} =1+1 =2\gt 1.$$ The value exceeds 1, so the point is outside the ellipse.

Only Option C satisfies the equation of the ellipse.

Hence, the correct answer is Option C.