1. Find the common tangents to the circle and parabola

Let the tangent be of slope form $y=mx+c.$

For the parabola $y^2=4x$, the tangent in slope form is $y=mx+\frac{1}{m}.$ So for a line tangent to the parabola, we must have $c=\frac{1}{m}.$

Now this same line must also be tangent to the circle $x^2+y^2=\frac12,$ which has centre $(0,0)$ and radius $r=\frac{1}{\sqrt2}.$

The distance of the line $mx-y+\frac{1}{m}=0$ from the origin must equal $r$: $\frac{\left|\frac1m\right|}{\sqrt{m^2+1}}=\frac{1}{\sqrt2}.$

Squaring, $\frac{1}{m^2(m^2+1)}=\frac12.$ So, $2=m^2(m^2+1).$ Let $t=m^2$. Then $t^2+t-2=0$ $\Rightarrow (t-1)(t+2)=0.$ Since $t\ge 0$, we get $m^2=1 \Rightarrow m=\pm 1.$

Hence the two common tangents are $y=x+1, \qquad y=-x-1.$

2. Find their point of intersection $Q$

Solve $y=x+1, \qquad y=-x-1.$ Then $x+1=-x-1 \Rightarrow 2x=-2 \Rightarrow x=-1,$ and $y=0.$ Thus, $Q=(-1,0).$

Therefore, $OQ=1.$

3. Form the ellipse

The ellipse is centred at the origin, and its semi-major axis is $a=OQ=1.$

The minor axis length is given as $\sqrt2$, so the semi-minor axis is $2b=\sqrt2 \Rightarrow b=\frac{1}{\sqrt2}.$ Thus, $b^2=\frac12.$

Hence the ellipse is $\frac{x^2}{1^2}+\frac{y^2}{(1/\sqrt2)^2}=1$ $x^2+2y^2=1.$

4. Check eccentricity and latus rectum

For an ellipse, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac12}=\frac{1}{\sqrt2}.$ This is also written as $\frac{\sqrt2}{2}.$

Length of latus rectum: $\ell=\frac{2b^2}{a}=\frac{2\cdot \frac12}{1}=1.$

So option A states eccentricity $\frac{1}{\sqrt2}$ and latus rectum $1$, which is correct.

Option B is false.

5. Area bounded by the ellipse between $x=\frac{1}{\sqrt2}$ and $x=1$

From $x^2+2y^2=1,$ we get $y=\pm \sqrt{\frac{1-x^2}{2}}=\pm \frac{1}{\sqrt2}\sqrt{1-x^2}.$

So the required area is $A=\int_{1/\sqrt2}^{1}\left(\text{upper }y - \text{lower }y\right)dx$ $=\int_{1/\sqrt2}^{1} 2\cdot \frac{1}{\sqrt2}\sqrt{1-x^2}\,dx$ $=\sqrt2\int_{1/\sqrt2}^{1}\sqrt{1-x^2}\,dx.$

Use $\int \sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x.$

Therefore, $A=\sqrt2\left[\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x\right]_{1/\sqrt2}^{1}.$

At $x=1$: $\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x=0+\frac12\cdot \frac\pi2=\frac\pi4.$

At $x=\frac{1}{\sqrt2}$: $\frac{x}{2}\sqrt{1-x^2}=\frac{1/\sqrt2}{2}\cdot \frac{1}{\sqrt2}=\frac14,$ $\frac12\sin^{-1}\left(\frac{1}{\sqrt2}\right)=\frac12\cdot \frac\pi4=\frac\pi8.$ So total value is $\frac14+\frac\pi8.$

Hence,