Area of a triangle inside an ellipse

Question

$F_1$, $F_2$ are are foci of the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$. $P$ is a point on the ellipse such that $|PF_1|:|PF_2|=2:1\;$, then how could I figure out of the area of $∆PF_1F_2$?

As we know $c^2=a^2-b^2=9-4=5$,
$$∴c=\pm \sqrt5$$ So the epicenter of the ellipse respectively $(\sqrt5,0)\; \& \;(-\sqrt5,0)$.
We've to find the area of $∆PF_1F_2$ which is $=1/2\times F_1F_2\times$(perpendicular distance from $P$ to any point of the horizontal line $F_1F_2$)
I'm not understanding what & how to do next... find out the area of ∆PF₁F₂

Answer 1

Hint:

WLOG $P(3\cos t,2\sin t)$

$$|PF_1|^2=(3\cos t-\sqrt5)^2+(2\sin t-0)^2=?$$

$$|PF_2|^2=(3\cos t+\sqrt5)^2+(2\sin t-0)^2=?$$

$$\dfrac{|PF_1|^2}{|PF_2|^2}=2^2$$

Use $\sin^2t=1-\cos^2t$ to form a Quadratic Equation in $\cos t$

Answer 2

Referring to the graph:

$\hspace{2cm}$

Note that $|PF_1|+|PF_2|=|AF_1|+|AF_2|=2a=6$. Using the given condition $|PF_1|=2|PF_2|$, we find $|PF_1|=4, |PF_2|=2$. You can use Heron's formula to find the area of $\Delta PF_1F_2$: $$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2\sqrt{5}, p=\frac{a+b+c}{2}=3+\sqrt{5}; \\ S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{(3+\sqrt{5})(\sqrt{5}-1)(\sqrt{5}+1)(3-\sqrt{5})}=4.$$

Answer 3

With the use of the definition of ellipse $$|PF_1|+|PF_2|=2a=6$$ and the given ratio $$|PF_1|:|PF_2|=2:1,$$ we get $$|PF_1|=4, \; |PF_2|=2$$ Since $F_1F_2=2\sqrt 5=\sqrt{20}=\sqrt{4^2+2^2},$ we have a right triangle $\triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$\mathcal{A}=\frac 12 \cdot |PF_1|\cdot |PF_2|=4$$