Consider two straight lines, each of which is tangent to both the circle $x^{2} + y^{2} = \frac{1}{2}$ and the parabola $y^{2} = 4 x$
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 the which of the following statement(s) is (are) TRUE?

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For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is 1

For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$

The area of the region bounded by the ellipse between the lines $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is $\frac{1}{4 \sqrt{2}} (π - 2)$

The area of the region bounded by the ellipse between the lines $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is $\frac{1}{16} (π - 2)$

answer is A, C.

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Detailed Solution

Let equation of common tangent is $y = m x + \frac{1}{m}$

$|\frac{0 + 0 + \frac{1}{m}}{\sqrt{1 + m^{2}}}| = \frac{1}{\sqrt{2}} \Rightarrow m^{4} + m^{2} - 2 = 0 \Rightarrow m = \pm 1$

Equation of common tangents are $y = x + 1$ and $y = - x - 1$

Point Q is $(- 1, 0)$

Equation of ellipse is $\frac{x^{2}}{1} + \frac{y^{2}}{1 / 2} = 1$

A) $e = \sqrt{1 + \frac{1}{2}} = \frac{1}{2} and L R = \frac{2 b^{2}}{a} = 1$

$Area 2. \begin{array}{r} \int_{1 / \sqrt{2}}^{2} \frac{1}{\sqrt{2}} \cdot \sqrt{1 + x^{2}} d x = \sqrt{2} {[\frac{x}{2} \sqrt{1 - x^{2}} 9 \frac{1}{2} \sin^{- 1} x]}_{1 \sqrt{2}}^{1} \\ = \sqrt{2} [\frac{π}{4} - (\frac{1}{4} + \frac{π}{8})] = \sqrt{2} (\frac{π}{8} - \frac{1}{4}) = \frac{π - 2}{4 \sqrt{2}} \end{array}$