If $\frac{(3x-4y-1)^2}{100}-\frac{(4x+3y-1)^2}{225}=1$, then find the length of the latus rectum > If $$\frac{(3x-4y-1)^2}{100}-\frac{(4x+3y-1)^2}{225}=1$$ then find the length of the latus rectum. If the standard hyperbola $\frac{x^2}{a^2}−\frac{y^2}{b^2}=1$, then the latus rectum is $2b^2/a$, but I am not able to apply the concept to the inclined hyperbola.

Rewrite the equation as,

$$\frac{\left( \frac35 x-\frac45 y-\frac15\right)^2}{4}-\frac{\left( \frac45 x+\frac35 y-\frac15\right)^2}{9}=1$$

which is the regular hyperbola

$$\frac{\left( x-\frac15\right)^2}{4}-\frac{\left( y-\frac15\right)^2}{9}=1$$

with the rotation of $\theta=\cos^{-1}\frac35$. Therefore,

$$a=2, \>\>\>\>\>b=3$$

which allows the latus rectum $2b^2/a$ to be calculated.