Length of Latus Rectum of $\frac{(x-y-2)^2}{16}-\frac{(x+y-4)^2}{9}=1$

Question

Find Length of Latus Rectum of Hyperbola $$\frac{(x-y-2)^2}{16}-\frac{(x+y-4)^2}{9}=1$$

we have Hyperbola as

$$\frac{\left(\frac{x-y-2}{\sqrt{2}}\right)^2}{8}-\frac{\left(\frac{x+y-4}{\sqrt{2}}\right)^2}{4.5}=1$$

Whose Transverse axis is $x+y=4$ and Conjugate axis is $x-y-2=0$

Comparing with standard Hyperbola Latus rectum length is $\frac{2b^2}{a}=\frac{9}{2\sqrt{2}}$

But the answer is $\frac{9}{2}$, Any thing wrong here?

Answer 1

GeoGebra confirms your result: