Analytic Geometry – Problem 21: find the perimeter of

Given the ellipse $C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\,(a>b>0)$. Let $A$ be the upper endpoint of its minor axis, and $F_1,F_2$ its foci. The eccentricity is $e=\frac12$. The line through $F_1$ perpendicular to $AF_2$ meets the ellipse at $D$ and $E$. If $|DE|=6$, find the perimeter of $\triangle ADE$.

A. 11 B. 12 C. 13 D. 14

Place the ellipse in standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $F_1(-c,0),F_2(c,0)$ and upper minor-axis vertex $A(0,b)$.

Since $e=\frac{c}{a}=\frac12$, we have $a=2c$. Then $$AF_1=AF_2=\sqrt{b^2+c^2}=\sqrt{a^2}=a=2c,$$ so $AF_1=AF_2=F_1F_2$. Hence $\triangle AF_1F_2$ is equilateral, and the line through $F_1$ perpendicular to $AF_2$ is the perpendicular bisector of segment $AF_2$. Therefore reflection across line $DE$ maps $A$ to $F_2$, and keeps $D,E$ fixed, so $$\text{perimeter}(\triangle ADE)=\text{perimeter}(\triangle F_2DE).$$

The perpendicular to $AF_2$ has slope $\frac{\sqrt3}{3}$, so line $DE$ has equation $$y=\frac{\sqrt3}{3}(x+c).$$ With $a=2c$ and $b^2=a^2-c^2=3c^2$, the ellipse is $$\frac{x^2}{4c^2}+\frac{y^2}{3c^2}=1.$$ Substitute the line into the ellipse to get a quadratic in $x$; the chord length along a line of slope $\frac{\sqrt3}{3}$ simplifies to $$|DE|=\frac{48}{13}c.$$ Given $|DE|=6$, we obtain $c=\frac{13}{8}$, hence $a=2c=\frac{13}{4}$ and $4a=13$.

Since $F_1$ lies on chord $DE$, we have $|F_1D|+|F_1E|=|DE|=6$. For each intersection point, ellipse definition gives $$|F_1D|+|F_2D|=2a,\qquad |F_1E|+|F_2E|=2a.$$ Add them: $$(|F_1D|+|F_1E|)+(|F_2D|+|F_2E|)=4a.$$ So $|F_2D|+|F_2E|=4a-|DE|=13-6=7$, and therefore $$\text{perimeter}(\triangle F_2DE)=(|F_2D|+|F_2E|)+|DE|=7+6=13.$$ Thus $\text{perimeter}(\triangle ADE)=13$, i.e. option C.

The perimeter is $13$ (option C).