Consider two straight lines, each of which is tangent to both the circle $x^{2} + y^{2} = \frac{1}{2}$ and the parabola $y^{2} = 4x$. Let these lines intersect at the point $Q$. Consider the ellipse whose center is at the origin $O(0,0)$ and whose semi-major axis is $OQ$. If the length of the minor axis of this ellipse is $\sqrt{2}$, then which of the following statement(s) is(are) TRUE?

$\text{Two lines tangent to both } x^2 + y^2 = \frac{1}{2} \text{ and } y^2 = 4x$ $\text{They intersect at } Q$ $\text{Ellipse centre } O(0,0), \text{ semi-major axis } = OQ$ $\text{Minor axis of the ellipse} = \sqrt{2}$ $\text{Which statements are true?}$ $x^2 + y^2 = \frac{1}{2}: \ \text{radius } \frac{1}{\sqrt{2}}$ $y^2 = 4x: \quad a = 1$ $\text{Tangent to } y^2 = 4x: \quad y = mx + \frac{1}{m}$ $d(O, \text{line}) = \frac{1}{|m|\sqrt{m^2+1}} = \frac{1}{\sqrt{2}}$ $m^2(m^2+1) = 2 \ \Rightarrow \ m^4 + m^2 - 2 = 0$ $(m^2+2)(m^2-1) = 0 \ \Rightarrow \ m^2 = 1$ $m = \pm 1$ $L_1: y = x + 1, \qquad L_2: y = -x - 1$ $Q = L_1 \cap L_2 = (-1, 0)$ $OQ = \sqrt{(-1)^2 + 0^2} = 1$ $\text{semi-major axis } a = OQ = 1$ $\text{minor axis} = \sqrt{2} \Rightarrow b = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ $\frac{x^2}{1^2} + \frac{y^2}{b^2} = 1$ $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$ $\text{Latus rectum} = \frac{2b^2}{a} = \frac{2 \cdot \frac{1}{2}}{1} = 1$ $x^2 + 2y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{2}}\sqrt{1-x^2}$ $A = \int_{1/\sqrt{2}}^{1} 2\cdot\frac{1}{\sqrt{2}}\sqrt{1-x^2}\,dx = \sqrt{2}\int_{1/\sqrt{2}}^{1}\sqrt{1-x^2}\,dx$ $\int \sqrt{1-x^2}\,dx = \frac{x\sqrt{1-x^2} + \sin^{-1}x}{2}$ $\int_{1/\sqrt{2}}^{1}\sqrt{1-x^2}\,dx = \frac{\pi}{4} - \left(\frac{1}{4} + \frac{\pi}{8}\right) = \frac{\pi}{8} - \frac{1}{4}$ $A = \sqrt{2}\left(\frac{\pi}{8} - \frac{1}{4}\right) = \frac{1}{4\sqrt{2}}(\pi - 2)$ $\therefore \text{Statements (1) and (3) are TRUE}$