Question

Find the equation of the hyperbola whose foci are (8,3),(0,3) and eccentricity =$4/3$.

Answer 1

The center of hyperbola is the midpoint of the line joining the two focii. So the coordinate of the centre are $(\frac{8+0}{2},\frac{3+3}{2})=(4,3)$

Let $2a$ and $2b$ be the length of transverse and conjugate axis and let $e$ be eccentricity. Then equality of hyperbola

$\frac{{(x-4)}^2}{a^2}-\frac{{(y-3)}^2}{b^2}=1$

Distance between focii $=2ae$

$ac=4$ $a=3$

$b^2=a^2(e^2-1)$

$b^2=9(-1+\frac{16}{9})=7$

$\frac{{(x-4)}^2}{9}-\frac{{(y-3)}^2}{7}=1$

${7x}^2-{9y}^2-56x+54y-32=0$