Analytic Geometry – Problem 15: find fixed point

Given $C_1:\dfrac{x^2}{4}-y^2=1$, $C_2:\dfrac{x^2}{4}+y^2=1$, lines through $P(2,1)$ meet them at $A,B,M,N$. If $k_{QA}+k_{QB}+k_{QM}+k_{QN}$ is constant, find fixed point $Q$.

Let $Q=(h,k_Q)$. Parameterize a line through $P(2,1)$ as $$x=2+t\cos\theta,\qquad y=1+t\sin\theta\quad(\tan\theta=\lambda).$$ Substitute into $C_1$: $$(\cos^2\theta-4\sin^2\theta)t^2+4(\cos\theta-2\sin\theta)t-4=0.$$ So for intersections $A,B$, Vieta gives $$t_A+t_B=-\frac{4(\cos\theta-2\sin\theta)}{\cos^2\theta-4\sin^2\theta},\qquad t_At_B=-\frac{4}{\cos^2\theta-4\sin^2\theta}.$$ Similarly, for $C_2$: $$(\cos^2\theta+4\sin^2\theta)t^2+4(\cos\theta+2\sin\theta)t+4=0,$$ and Vieta gives $t_M+t_N, t_Mt_N$.

Now $$k_{QA}=\frac{1+t_A\sin\theta-k_Q}{2+t_A\cos\theta-h},\quad k_{QB}=\frac{1+t_B\sin\theta-k_Q}{2+t_B\cos\theta-h},$$ and analogously for $k_{QM},k_{QN}$. Using the Vieta sums/products, both $k_{QA}+k_{QB}$ and $k_{QM}+k_{QN}$ become rational functions of $\tan\theta$ with coefficients depending on $h,k_Q$. Therefore $$S(\theta)=k_{QA}+k_{QB}+k_{QM}+k_{QN}=A_0+A_1\tan\theta+A_2\tan^2\theta.$$ For $S(\theta)$ to be constant for all directions, coefficients of $\tan\theta$ and $\tan^2\theta$ must vanish, giving $$h=2,\qquad k_Q=0.$$ Hence the fixed point is $$Q=(2,0).$$

$$Q=(2,0).$$