Question

The length of the transverse axis of the hyperbola $9x^2-16y^2-18x-32y-151=0$ is

Answer 1

Answer: 8

The equation of given hyperbola is
$9x^2-16y^2-18x-32y-151=0$
We can simplify it as
$9x^2-18x+9-16y^2-32y-16-151+7=0$
$\Rightarrow 9{\left(x-1\right)}^2-16{\left(y+1\right)}^2=144$
$\Rightarrow \frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{9}=1$
Length of the transverse axis is $2a=8.$