If tangents are drawn to the ellipse $$x^2 + 2y^2 = 2$$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve:

We have the ellipse $$x^{2}+2y^{2}=2$$.

First divide every term by $$2$$ so that the equation attains the usual standard form:

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

By comparison with the general ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ we identify $$a^{2}=2$$ and $$b^{2}=1$$.

For an ellipse, the tangent in slope form is given by the well-known result

$$y=mx\;\pm\;\sqrt{a^{2}m^{2}+b^{2}},$$

because, for the line $$y=mx+c$$ to touch the ellipse, the condition $$c^{2}=a^{2}m^{2}+b^{2}$$ must hold (discriminant of the quadratic obtained on substituting vanishes).

Substituting $$a^{2}=2,\;b^{2}=1,$$ the equation of any tangent (except at a vertex where the slope is infinite or zero) becomes

$$y=mx\;\pm\;\sqrt{2m^{2}+1}.$$

Let us write this line as $$y=mx+c$$ with

$$c=\pm\sqrt{2m^{2}+1}.$$

Now we find its intercepts on the coordinate axes.

• On the y-axis put $$x=0$$:

$$y=c.$$

Hence the y-intercept is $$B\,(0,c).$$

• On the x-axis put $$y=0$$:

$$0=mx+c\;\;\Longrightarrow\;\;x=-\dfrac{c}{m}.$$

So the x-intercept is $$A\!\left(-\dfrac{c}{m},\,0\right).$$

We are interested in the midpoint $$P(h,k)$$ of the segment $$AB$$ cut off between the axes. Using the midpoint formula,

$$\begin{aligned} h&=\frac{-\dfrac{c}{m}+0}{2}=-\frac{c}{2m},\\[6pt] k&=\frac{0+c}{2}=\frac{c}{2}. \end{aligned}$$

To obtain the locus we must eliminate the parameters $$m$$ and $$c$$. Start by expressing $$c$$ in terms of $$k$$ from the second relation:

$$c=2k.$$

Squaring both sides of the defining relation $$c^{2}=2m^{2}+1$$ gives

$$c^{2}=2m^{2}+1.$$

Substitute $$c=2k$$:

$$\left(2k\right)^{2}=2m^{2}+1\;\;\Longrightarrow\;\;4k^{2}=2m^{2}+1.$$

Solve this for $$m^{2}$$:

$$2m^{2}=4k^{2}-1\;\;\Longrightarrow\;\;m^{2}=\frac{4k^{2}-1}{2}.$$

Next square the expression for $$h$$:

$$h^{2}=\left(-\frac{c}{2m}\right)^{2}=\frac{c^{2}}{4m^{2}}.$$

Insert $$c^{2}=4k^{2}$$ and $$m^{2}=\dfrac{4k^{2}-1}{2}:$$

$$h^{2}=\frac{4k^{2}}{4\left(\dfrac{4k^{2}-1}{2}\right)} =\frac{4k^{2}}{2\,(4k^{2}-1)} =\frac{2k^{2}}{4k^{2}-1}.$$

To bring $$h$$ and $$k$$ into one relation, cross-multiply:

$$h^{2}\,(4k^{2}-1)=2k^{2}.$$

Expand and rearrange every term to one side:

$$4h^{2}k^{2}-h^{2}-2k^{2}=0.$$

Now isolate the constant $$1$$ by dividing through by $$4h^{2}k^{2}$$ (remember $$h,k\neq0$$ since the points of tangency are not vertices):

$$1=\frac{h^{2}}{4h^{2}k^{2}}+\frac{2k^{2}}{4h^{2}k^{2}} =\frac{1}{4k^{2}}+\frac{1}{2h^{2}}.$$

Interchanging $$h$$ with $$x$$ and $$k$$ with $$y$$ to express the locus in the usual $$xy$$-plane, we finally get

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

This is exactly the equation offered in option C.

Hence, the correct answer is Option C.