Let the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and $y^{2} = 4 x$ intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6 , respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to

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Detailed Solution

$x^{2} + y^{2} = \frac{9}{4}, y = 4 x Equation tangent in slope form \begin{aligned} y = mx \pm \frac{3}{2} \sqrt{(1 + m^{2})} - - - (1) \\ y = mx + \frac{1}{m} - - - - (2) compare (1) \& (2) \\ \begin{array}{l} \pm \frac{3}{2} \sqrt{(1 + m^{2})} = \frac{1}{m^{}} \\ 9 m^{2} (1 + m^{2}) = 4 \\ 9 m^{4} + 9 m^{2} - 4 = 0 \\ 9 m^{4} + 12 m^{2} - 3 m^{2} - 4 = 0 \\ 3 m^{2} (3 m^{2} + 4) - 1 (3 m^{2} + 4) = 0 \end{array} \end{aligned}$

$\begin{aligned} m^{2} = - \frac{4}{3} (Rejected) \\ m^{2} = \frac{1}{3} \Rightarrow m = \pm \frac{1}{\sqrt{3}} \end{aligned}$

Equation of common tangent is

$y = \frac{1}{\sqrt{3}} x + \sqrt{3}$

$\begin{array}{l} OQ = - 3 \\ b = | OQ | = 3 \\ a = 6 \\ b^{2} = a^{2} (1 - e^{2}) \Rightarrow e^{2} = 1 - \frac{9}{36} = \frac{3}{4} \\ l = \frac{2 b^{2}}{a} = \frac{2 \times 9}{6} = 3 \\ \frac{l}{e^{2}} = \frac{3}{3 / 4} = 4 \end{array}$