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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 which of the following statement($s$) is (are) $TRUE$?
$(A)$ For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $1$
$(B)$ For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$
$(C)$ The area of the region bounded by the ellipse between the lines $x=\frac{1}{\sqrt{2}}$ and $x=1$ is $\frac{1}{4 \sqrt{2}}(\pi-2)$
$(D)$ The area of the region bounded by the ellipse between the lines $x=\frac{1}{\sqrt{2}}$ and $x=1$ is $\frac{1}{16}(\pi-2)$
$A,B$
$A,D$
$A,C$
$A,B,C$
Solution
Eq. of tangent to $y^2=4 x$ is
$y = mx +\frac{1}{ m }$
Eq. of tangent to $x^2+y^2=\left(\frac{1}{\sqrt{2}}\right)^2$ is
$y=m x \pm \frac{\sqrt{1+m^2}}{\sqrt{2}}$
Comparing (1) and (2), we get $m ^2=1 \Rightarrow m = \pm 1$
$\therefore Q \equiv(-1,0)$
$\therefore$ Eq. of ellipse is $\frac{x^2}{1^2}+\frac{y^2}{\left(\frac{1}{2}\right)}=1$
Eccentricity $=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
Length of latus rectum $=2 \cdot \frac{1}{2}=1$
$\text { Area } =2 \int_{\frac{1}{\sqrt{2}}}^1 \sqrt{\frac{1-x^2}{2}} d x$
$=\sqrt{2}\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}(x)\right]_{\frac{1}{\sqrt{2}}}^1=\frac{(\pi-2)}{4 \sqrt{2}}$
