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(C) The equation of a tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.The equation of a tangent to the circle $x^2 + y^2 = \frac{1}{2}$ i

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$A, C$

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(C) The equation of a tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.The equation of a tangent to the circle $x^2 + y^2 = \frac{1}{2}$ is $y = mx \pm \sqrt{\frac{1}{2}(1 + m^2)}$.Comparing the two,we have $\frac{1}{m^2} = \frac{1 + m^2}{2} \Rightarrow 2 = m^2 + m^4 \Rightarrow m^4 + m^2 - 2 = 0$.$(m^2 + 2)(m^2 - 1) = 0$,so $m^2 = 1$,which gives $m = \pm 1$.The tangents are $y = x + 1$ and $y = -x - 1$. Their intersection point $Q$ is $(-1, 0)$.The distance $OQ = 1$,which is the semi-major axis $a = 1$.The length of the minor axis is $2b = \sqrt{2}$,so $b = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,which means $b^2 = \frac{1}{2}$.The equation of the ellipse is $\frac{x^2}{1} + \frac{y^2}{1/2} = 1$.Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/2}{1}} = \frac{1}{\sqrt{2}}$.Length of latus rectum $= \frac{2b^2}{a} = \frac{2(1/2)}{1} = 1$. Thus,$(A)$ is true.Area $= 2 \int_{1/\sqrt{2}}^{1} \sqrt{\frac{1}{2}(1 - x^2)} dx = \sqrt{2} \int_{1/\sqrt{2}}^{1} \sqrt{1 - x^2} dx$.$= \sqrt{2} \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}(x) \right]_{1/\sqrt{2}}^{1} = \sqrt{2} \left( (0 + \frac{1}{2} \cdot \frac{\pi}{2}) - (\frac{1}{2\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{\pi}{4}) \right)$.$= \sqrt{2} \left( \frac{\pi}{4} - \frac{1}{4} - \frac{\pi}{8} \right) = \sqrt{2} \left( \frac{\pi}{8} - \frac{1}{4} \right) = \frac{\sqrt{2}}{8}(\pi - 2) = \frac{1}{4\sqrt{2}}(\pi - 2)$. Thus,$(C)$ is true.

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