Let the common tangents to the curves 4(x^2+y^2)=9 and y^2=4x intersect at the point Q.

Question

Let the common tangents to the curves $4(x^2+y^2)=9$ and $y^2=4x$ intersect at the point $Q$. Let an ellipse, centred at the origin $O$, have 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 $\dfrac l{e^2}$ is equal to\_\_\_\_

Answer: 4

Accepted Answer

Circle: $x^2+y^2=\tfrac94$. Tangent to $y^2=4x$: $y=mx+1/m$; tangency to the circle gives $9m^4+9m^2-4=0\Rightarrow m^2=\tfrac13$.

The two common tangents ($m=\pm1/\sqrt3$) meet at $Q=(-3,0)$ (by symmetry, on the $x$-axis), so $OQ=3$.

Ellipse: semi-minor $b=3$, semi-major $a=6$: $e^2=1-\dfrac9{36}=\dfrac34$, $l=\dfrac{2b^2}a=3$.

$$\frac l{e^2}=\frac3{3/4}=4.$$

Solutions · 1