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.
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HINT.
The transformation $$ x'=3x-4y\\ y'=4x+3y\\ $$ is not only a rotation, but also a dilation: try it on some points to see how it works. You should re-define it to avoid the dilation.
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Promoting my comment to an answer ...
Note that $3x-4y-1=0$ and $4x+3y-1=0$ are perpendicular lines (why?), and recall that the distance from a point $(x,y)$ to the line $px+qy-r=0$ is $$d = \frac{|px+qy-r|}{\sqrt{p^2+q^2}} \tag{1}$$ Thus, the distances from $(x,y)$ to the perpendicular lines are $$u := \frac15|3x-4y-1| \qquad v:=\frac15|4x+3y-1| \tag{2}$$
The given hyperbola equation then can be re-written as
$$\frac{u^2}{4}-\frac{v^2}{9}=1 \tag{3}$$
Since the lines are perpendicular, we recognize $(3)$ as the standard equation of the hyperbola that uses them as axes. From here, computing the latus rectum is straightforward. $\square$
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.