Krdytkyu

$F_1$, $F_2$ 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 $=1/2times F_1F_2times$(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

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 $Delta PF_1F_2$:
$$ a=|PF_1|=4, b=|PF_2|=2, c=|F_1F_2|=2sqrt{5}, p=frac{a+b+c}{2}=3+sqrt{5}; \
S=sqrt{p(p-a)(p-b)(p-c)}=sqrt{(3+sqrt{5})(sqrt{5}-1)(sqrt{5}+1)(3-sqrt{5})}=4.$$

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 $F_1F_2=2sqrt 5=sqrt{20}=sqrt{4^2+2^2},$ we have a right triangle $triangle PF_1F_2$ with hypotenuse $F_1F_2.$ The area is $$mathcal{A}=frac 12 cdot |PF_1|cdot |PF_2|=4$$

Hint:

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

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

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

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

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