Question

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________.

Answer 1

$x ^{2}+ y ^{2}=\frac{9}{4}\,\,\,\,\, y =4 x$
Equation tangent in slope form
$y=m x+\frac{3}{2} \sqrt{\left(1+m^{2}\right)} $
$y=m x+\frac{1}{m} $
compare (1) & (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 $
$3 m^{2}\left(3 m^{2}+4\right)-\left(3 m^{2}+4\right)=0$
$m^{2}=-\frac{4}{3}$ ( Rejected )
$m^{2}=\frac{1}{3} \Rightarrow m =\pm \frac{1}{\sqrt{3}}$
Equation of common tangent
$y=\frac{1}{\sqrt{3}} x+\sqrt{3}$
on $X$ axis $y =0$
$OQ =-3$
$b =| OQ |=3$
$a =6$
$b ^{2}= a ^{2}\left(1- e ^{2}\right) $
$\Rightarrow e ^{2}=1-\frac{9}{36}=\frac{3}{4}$
$e =\frac{2 b ^{2}}{ a }=\frac{2 \times 9}{6}=3$
$\frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$