Analytic Geometry – Problem 24: find the slope of

For the ellipse $E:\frac{x^2}{4}+\frac{y^2}{3}=1$, a line $l$ through the left focus $F_1$ meets the ellipse at $A$ and $B$ (with $A$ above the $x$-axis). If $$\frac{AF_1}{F_1B}=2,$$ find the slope $k$ of $l$.

A. $\frac{\sqrt3}{2}$ B. $\pm\frac{\sqrt3}{2}$ C. $\frac{\sqrt5}{2}$ D. $\pm\frac{\sqrt5}{2}$

For $E:\frac{x^2}{4}+\frac{y^2}{3}=1$, we have $a=2$, $b=\sqrt3$, so $c=\sqrt{a^2-b^2}=1$ and eccentricity $e=\frac{c}{a}=\frac12$.

Let $\lambda=\frac{AF_1}{F_1B}=2$. For a chord through a focus, the standard focal-radius ratio formula is $$e\cos\theta=\left|\frac{\lambda-1}{\lambda+1}\right|,$$ where $\theta$ is the inclination angle of the chord (so $k=\tan\theta$).

Compute: $$\left|\frac{\lambda-1}{\lambda+1}\right|=\left|\frac{2-1}{2+1}\right|=\frac13.$$ Thus $$\cos\theta=\frac{1/3}{e}=\frac{1/3}{1/2}=\frac23.$$ So $$\tan\theta=\frac{\sqrt{1-\cos^2\theta}}{\cos\theta}=\frac{\sqrt{1-4/9}}{2/3}=\frac{\sqrt{5}/3}{2/3}=\frac{\sqrt5}{2}.$$ Therefore $k=\pm\frac{\sqrt5}{2}$, i.e. option D.

$k=\pm\frac{\sqrt5}{2}$ (option D).