Answer
(a) $a=3, b=4$, $c=5$. $F_1(-5,0)$, $F_2(5,0)$
(b) see explanation.
(c) $d(P,F_1)=\frac{34}{3}$, $d(P,F_2)=\frac{16}{3}$
(d) see explanation.
Work Step by Step
(a) Given the hyperbola $\frac{x^2}{3^2} -\frac{y^2}{4^2}=1$, we can determine $a=3, b=4$ and $c=\sqrt {3^2+4^2}=5$. The transverse axis is horizontal and the foci locations are $F_1(-5,0)$ and $F_2(5,0)$
(b) Plug the coordinates of $P(5,\frac{16}{3})$ in the equation, we have $\frac{5^2}{3^2} -\frac{(\frac{16}{3})^2}{4^2}=\frac{25}{9} -\frac{16^2}{3^24^2}=\frac{25}{9} -\frac{16}{9}=1$, thus point P is on the hyperbola.
(c) Use the distance formula $d(P,F_1)=\sqrt {(5+5)^2+(16/3)^2}=\sqrt {100+256/9}=\frac{34}{3}$, and
$d(P,F_2)=\sqrt {(5-5)^2+(16/3)^2}=\sqrt {0+256/9}=\frac{16}{3}$
(d) The difference of $d's$ is $|d(P,F_1)-d(P,F_2)|=\frac{34-16}{3}=6=2a$ (as $a=3$)
