Analytic Geometry – Problem 27: Find the perimeter of the focal triangle

Let $E$ be the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ with foci $F_1(-3,0)$ and $F_2(3,0)$. Point $P$ is any point on $E$.

Find the perimeter of the focal triangle $\triangle F_1PF_2$.

(1) From $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, we have $a^2=25\Rightarrow a=5$, $b^2=16\Rightarrow b=4$.

(2) For an ellipse, $c^2=a^2-b^2$. Hence $$c=\sqrt{25-16}=3,$$ so $|F_1F_2|=2c=6$.

(3) By the ellipse definition, for any $P\in E$, $$|PF_1|+|PF_2|=2a=10.$$ Therefore the perimeter is $$|PF_1|+|PF_2|+|F_1F_2|=10+6=16.$$

The perimeter of $\triangle F_1PF_2$ is constant and equals $16$.