Analytic Geometry – Problem 23: find

Let the ellipse be $$\frac{x^2}{4}+y^2=1.$$ The upper and lower endpoints of its minor axis are $A(0,1)$ and $B(0,-1)$. A line $l$ given by $y=k(x-1)$ intersects the ellipse at points $C(x_1,y_1)$ and $D(x_2,y_2)$. Let the slopes of lines $AD$ and $CB$ be $k_1$ and $k_2$, respectively. If $k_1:k_2=2:1$, find $k$.

Because $A(0,1)$, $B(0,-1)$, we have $$k_1=\frac{y_2-1}{x_2},\qquad k_2=\frac{y_1+1}{x_1}.$$ Since $C,D\in l$, $y_i=k(x_i-1)$. Hence $$k_1=\frac{k(x_2-1)-1}{x_2},\qquad k_2=\frac{k(x_1-1)+1}{x_1}.$$ The condition $k_1:k_2=2:1$ gives $$\frac{k(x_2-1)-1}{x_2}=2\cdot\frac{k(x_1-1)+1}{x_1}.\tag{1}$$

Now find $x_1,x_2$. Substitute $y=k(x-1)$ into the ellipse: $$\frac{x^2}{4}+k^2(x-1)^2=1$$ which simplifies to $$(1+4k^2)x^2-8k^2x+4(k^2-1)=0.$$ Thus $x_1,x_2$ are roots of this quadratic. Eliminating $x_1,x_2$ between the Vieta relations and equation (1) yields $$(k-1)(k+1)(3k-1)^2=0.$$ So $k\in\{-1,\,\tfrac13,\,1\}$.

If $k=1$, then $l:y=x-1$ passes through $B(0,-1)$, making line $CB$ degenerate and $k_2$ not a well-defined slope. If $k=-1$, then $l:y=-(x-1)$ passes through $A(0,1)$, making $k_1$ ill-defined. Therefore the valid value is $$k=\frac13.$$

$$k=\frac13.$$