Question

Let the common tangents to the curves 4(x2 + y2) = 9 and y2 = 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 I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then
$\frac{1}{e^2}$ is equal to

Answer 1

The correct answer is 4
Let y = mx + c is the common tangent
So, $c=\frac{1}{m}=±\frac{3}{2}\sqrt{1+m^2}⇒m^2=\frac{1}{3}$
So equation of common tangents will be
$y=±\frac{1}{\sqrt3}x±\sqrt3$
which intersects at Q(–3, 0)
Major axis and minor axis of ellipse are 12 and 6. So eccentricity
$e^2=1−\frac{1}{4}=\frac{3}{4}$
and length of latus rectum $=\frac{2b^2}{a}=3$
Therefore ,
$\frac{l}{e^2}=\frac{3}{\frac{3}{4}}=4$