Left focus f? (-3,0); Right focus f? (3,0); A(0，6√6)； Find a small P in the left branch first, like this? APF？ The circumference is the smallest.
Geometric drawing: connecting AF? , and the left hyperbola intersect at p, and this intersection point p is? APF？ The point with the smallest circumference.
Proof: in? APF？ Medium, AF? It is fixed-length, so the following only needs to prove AP+PF? Keep it to a minimum.
Hyperbolic definition: PF? -PF？ =2a=2,∴PF？ =PF？ +2; ∴AP+PF？ =AP+PF？ +2=AF？ +2.
Take another p different from the p on the left branch? At this point, AP? +P？ f？ =AP？ +P？ f？ +2 & gt； AF？ +2; ? Is that AP? +P？ f？ & gtAP+PF？ ;
So this card.
Let's find the coordinates of point P: AF? K = 6 √ 6/3 = 2 √ 6 slope; So AF? The equation of is y = 2 (√ 6) (x+3);
Substitute into hyperbolic equation: x? -3(x+3)？ = 1,? Do you have an X? +9x+ 14=(x+2)(x+7)=0, then x? =-2，x？ =-7 (shed
Go); ? According to this, y? = √[ 8(x- 1)]= √[ 8(4- 1)]=√24 = 2√6； P (-2，2√6)；
At this time? APF？ Area:
Because of the website, the picture I drew became so small that I couldn't see it at all, so there was no picture.
Fortunately, this picture is easy to draw. I have expressed it clearly in words. Please draw it yourself.