Let the common tangents to the curves 4(x^{2} | vedclass

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(B) The circle is $x^{2} + y^{2} = \frac{9}{4}$ and the parabola is $y^{2} = 4x$.The tangent to the parabola $y^{2} = 4x$ is $y = mx + \frac{1}{m}$.Th

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$4$

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(B) The circle is $x^{2} + y^{2} = \frac{9}{4}$ and the parabola is $y^{2} = 4x$.The tangent to the parabola $y^{2} = 4x$ is $y = mx + \frac{1}{m}$.The distance from the origin $(0,0)$ to this tangent must equal the radius of the circle,which is $\frac{3}{2}$.$\frac{|m(0) - 0 + 1/m|}{\sqrt{m^{2} + 1}} = \frac{3}{2} \Rightarrow \frac{1}{|m|\sqrt{m^{2} + 1}} = \frac{3}{2}$.Squaring both sides: $\frac{1}{m^{2}(m^{2} + 1)} = \frac{9}{4} \Rightarrow 9m^{4} + 9m^{2} - 4 = 0$.Solving for $m^{2}$: $(3m^{2} - 1)(3m^{2} + 4) = 0$. Since $m^{2} > 0$,we have $m^{2} = \frac{1}{3}$,so $m = \pm \frac{1}{\sqrt{3}}$.The common tangents are $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ and $y = -\frac{1}{\sqrt{3}}x - \sqrt{3}$.These tangents intersect at the point $Q$ on the $x$-axis where $y=0$. Setting $y=0$ in $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ gives $x = -3$. Thus,$Q = (-3, 0)$.The length $OQ = |-3| = 3$. Given the ellipse has semi-minor axis $b = 3$ and semi-major axis $a = 6$.Eccentricity $e$ is given by $b^{2} = a^{2}(1 - e^{2})$ $\Rightarrow 9 = 36(1 - e^{2})$ $\Rightarrow 1 - e^{2} = \frac{1}{4}$ $\Rightarrow e^{2} = \frac{3}{4}$.Length of latus rectum $l = \frac{2b^{2}}{a} = \frac{2 	imes 9}{6} = 3$.Therefore,$\frac{l}{e^{2}} = \frac{3}{3/4} = 4$.

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