Analytic Geometry – Problem 6: Find the equation of ellipse

Ellipse $C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$ has eccentricity $\dfrac12$. Its left and right vertices are $A_1,A_2$, its left and right foci are $F_1,F_2$, and its upper vertex is $B$. The circumradius of $\triangle A_2BF_2$ is $\dfrac{2\sqrt{21}}3$.

(1) Find the equation of ellipse $C$.

(2) A variable line with finite slope intersects the ellipse at $P,Q$ (on opposite sides of the $x$-axis). Let the slopes of $PA_1,PA_2,QA_2,QA_1$ be $k_1,k_2,k_3,k_4$, and suppose $$k_1+k_4=\frac53(k_2+k_3).$$ Find the range of the area of $\triangle PQF_1$.

(1) From $e=\frac ca=\frac12$, we have $a=2c$, hence $b=\sqrt3c$. In right triangle $BF_2O$, $|BF_2|=a=2c$, $|OF_2|=c$. Together with the circumradius condition, $$\frac{|A_2B|}{\sin120^\circ}=2R=\frac{4\sqrt{21}}3.$$ we get $a^2+b^2=7$, so $c=1,a=2,b=\sqrt3$. Therefore $$C:\frac{x^2}{4}+\frac{y^2}{3}=1.$$

(2) Let $$PQ:y=k(x+m),\quad -2<m<2.$$ Substitute into the ellipse: $$(3+4k^2)x^2+8k^2mx+4k^2m^2-12=0.$$ Let intersections be $P(x_1,y_1),Q(x_2,y_2)$, and write $x_1+x_2,x_1x_2$ by Vieta.

Using the slope relation $$k_1+k_4=\frac53(k_2+k_3),$$ the official simplification gives $$2m^2+3m-2=0.$$ With $-2<m<2$, we obtain $$m=\frac12.$$ Hence the intersection of $PQ$ with the $x$-axis is fixed at $M(-\tfrac12,0)$, so $$|F_1M|=\frac12.$$

Let the area be $S$: $$S=\frac12|F_1M|\,|y_1-y_2|=\frac14|k|\,|x_1-x_2|.$$ Further simplification using Vieta gives $$0<S^2<\frac{45}{64},$$ therefore $$0<S<\frac{3\sqrt5}{8}.$$

(1) The ellipse equation is $$\frac{x^2}{4}+\frac{y^2}{3}=1.$$ It is determined by the eccentricity condition together with the given circumradius information.

(2) The slope relation forces $m=\frac12$, so the $x$-intercept of $PQ$ is fixed. Converting the area into a Vieta expression yields $$0<S<\frac{3\sqrt5}{8}.$$ Hence the area range of $\triangle PQF_1$ is $\left(0,\dfrac{3\sqrt5}{8}\right)$.