Question

The centre of the hyperbola $\frac{(3x-4y-12{)}^2}{225}-\frac{(4x+3y-12{)}^2}{100}=1$ is :

• A. $\left(\frac{-84}{25},\frac{12}{25}\right)$
• B. $\left(\frac{84}{25},\frac{-12}{25}\right)$
• C. $\left(\frac{84}{25},\frac{12}{25}\right)$
• D. $\left(\frac{-84}{25},\frac{-12}{25}\right)$

Answer 1

The correct option is B $\left(\frac{84}{25},\frac{-12}{25}\right)$

Centre of hyperbola is point of intersection of $3x-4y-12=0$ and $4x+3y-12=0$
$\Rightarrow 3x-4y=12$
$4x+3y=12$
Subtracting both equations
$\Rightarrow -x-7y=0$
$x=-7y$
$3(-7y)-4y=12$
$y=\frac{-12}{25}$
$x=\frac{84}{25}$