Answer:
Step-by-step explanation:
Remember, the points P on the hyperbola satisfy that the value absolute of the difference of the distances of P to the foci is constant and less than the distance between the foci.
Then
[tex]\lvert\lvert PF\lvert\lvert-\lvert\lvertPG\lvert\lvert=2a, \; \lvert\lvert PG\lvert\lvert-\lvert\lvert PF\lvert\lvert=2a[/tex]
Therefore, [tex]2a=6\\a=3[/tex]
Also, the foci [tex](0,c)=(0,5), \; (0,-c)=(0,5)[/tex] satisfy that [tex]c=\sqrt{a^2+b^2}[/tex], then
[tex]5=\sqrt{3^2+b^2}\\5^2=3^2+b^2\\25-9=b^2\\16=b^2[/tex]
Then, the equaton of the hyperbola is
[tex]\frac{x^2}{3^2}-\frac{y^2}{16}=1\\\frac{x^2}{9}-\frac{y^2}{16}=1[/tex]