Question

The equation of the hyperbola whose foci are $(8,3)$ and $(0,3)$ and eccentricity$=\frac{4}{3}$ is

• A. $7{(x-4)}^2-9{(y-3)}^2=63$
• B. ${7x}^2-{9y}^2=63$
• C. $9{(x-4)}^2-9{(y-3)}^2=63$
• D. $7{(x+4)}^2-9{(y+3)}^2=63$

Answer 1

The correct option is A $7{(x-4)}^2-9{(y-3)}^2=63$

The centre of the hyperbola is the mid-point of the line joining the two
foci. So, the coordinates of the centre are $(\frac{8+0}{2},\frac{3+3}{2})$ i.e $(4,3)$
Let $2a,2b$ be the length of the transverse and conjugate axes and let $e$ be the
eccentricity. Then, the equation of the hyperbola is
$\frac{{(x-4)}^2}{a^2}-\frac{{(y-3)}^2}{b^2}=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$
Distance between the two foci$=2ae$
$\Rightarrow \sqrt{{(8-0)}^2+{(3-3)}^2}=2ae$
$\Rightarrow ae=4\Rightarrow a=3$
$\therefore b^2=a^2(e^2-1)\Rightarrow b^2=9(\frac{16}{9}-1)=7$
Substituting the value of $a$ and $b$ in $\left(i\right)$, we find that the equation of the hyperbola is
$\frac{{(x-4)}^2}{9}+\frac{{(y-3)}^2}{7}=1or7{(x-4)}^2-9{(y-3)}^2=63$
Hence, option 'A' is correct.