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Let the common tangents to the curves $4\left(x^{2}+y^{2}\right)=$ $9$ 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
$5$
$4$
$3$
$2$
Solution
$x^{2}+y^{2}=\frac{9}{4} \quad y=4 x$
Equation tangent in slope form
$y=m x \pm \frac{3}{2} \sqrt{\left(1+m^{2}\right)}$ …..$(1)$
$y=m x+\frac{1}{m}$ …..$(2)$
compare $(1)$ and $(2)$
$\pm \frac{3}{2} \sqrt{\left(1+m^{2}\right)}=\frac{1}{m^{2}}$
$9 m^{2}\left(1+m^{2}\right)=4$
$9 m^{4}+9 m^{2}-4=0$
$9 m^{4}+12 m^{2}-3 m^{2}-4=0$
$3 m^{2}\left(3 m^{2}+4\right)-\left(3 m^{2}+4\right)=0$
$m^{2}=-\frac{4}{3}(\text { Rejected })$
$m^{2}=\frac{1}{3} \Rightarrow m=\pm \frac{1}{\sqrt{3}}$
Equation of common tangent
$y =\frac{1}{\sqrt{3}} x +\sqrt{3}$
$\text { on } X \text { axis } y =0$
$OQ =-3$
$b =\mid OQ\mid =3$
$a =6$
$b ^{2}= a ^{2}\left(1- e ^{2}\right) \Rightarrow e ^{2}=1-\frac{9}{36}=\frac{3}{4} ~\\ e =\frac{2 b ^{2}}{ a }=\frac{2 \times 9}{6}=3 ~\\ \frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$
$\frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$
