Question

what is the lenght of the conjugate axis?

(x-1)^2/25 -(y+3)^2/9=1

Answer 1

$\frac{(x-1)^2}{25}- \frac{(y+3)^2}{9}=1$ ........ (1)

The standard equation of the hyperbola is given by:

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

When we compare the standard equations and equation (1) we obtain

$a^2= 25$ and $b^2=9$

This implies that $a=5$ and $b=3$

Conjugate axis$=2b$

$=2 \times 3=6$ units