Question

Find the centre, the foci, the directrices, of the hyperbola $\frac{(x-3{)}^2}{9}-\frac{(y-4{)}^2}{16}=1$

Answer 1

The equation of hyperbola $\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1$

Given equation is $\frac{(x-3{)}^2}{9}-\frac{(y-4{)}^2}{16}=1$

Center is $(h,k)=(3,4)$

$e=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{9+16}{9}}=\frac{5}{3}$

Foci is $(h\pm ae,k)=(3\pm 3\frac{5}{3},4)=(3\pm 5,4)$

Directrix $x=h\pm a/ex=3\pm \frac{3}{\frac{5}{3}}x=3\pm \frac{9}{5}$