Question

Let the common tangents to the curves $4\left(x^2+y^2\right)=9$ and $y^2=4x$ 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

Answer: 4

$x^2+y^2=\frac{9}{4}y=4x$
Equation tangent in slope form
$y=mx+\frac{3}{2}\sqrt{\left(1+m^2\right)}$
$y=mx+\frac{1}{m}$
compare (1) & (2)
$\pm \frac{3}{2}\sqrt{\left(1+m^2\right)}=\frac{1}{m^2}$
$9m^2\left(1+m^2\right)=4$
$9m^4+9m^2-4=0$
$9m^4+12m^2-3m^2-4=0$
$3m^2\left(3m^2+4\right)-\left(3m^2+4\right)=0$
$m^2=-\frac{4}{3}$ ( 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 $y=0$
$OQ=-3$
$b=|OQ|=3$
$a=6$
$b^2=a^2\left(1-e^2\right)$
$\Rightarrow e^2=1-\frac{9}{36}=\frac{3}{4}$
$e=\frac{2b^2}{a}=\frac{2\times 9}{6}=3$
$\frac{e}{e^2}=\frac{3}{3/4}=4$