The equations of the transverse and conjugate axes of a hyperbola are respectively $\mathrm{x}+2\mathrm{y}-3=0, 2\mathrm{x}-\mathrm{y}+4=0$ and their respective lengths are $\sqrt{2}$ and $2/\sqrt{3}$. The equation of the hyperbola is

Given

   $$\begin{gathered} 2\mathrm{a}=\sqrt{2}, 2\mathrm{b}=\frac{2}{\sqrt{3}} \\ \Rightarrow a=\frac{1}{\sqrt{2}},b=\frac{1}{\sqrt{3}} \end{gathered}$$

If we  take the two axes as the new coordinate system, and point of intersection of the axes as the new origin, then in the new coordinae system, equation of hyperbola will be 

$$\begin{gathered} \frac{X^2}{a^2}-\frac{Y^2}{b^2}=1 \\ \Rightarrow 2X^2-3Y^2=1 \end{gathered}$$

 Let P(x,y) be the coordinates of a point on the hyperbola in original x-y system, then

$\mathrm{X}=\frac{2\mathrm{x}-\mathrm{y}+4}{\sqrt{5}},\mathrm{Y}=\frac{\mathrm{x}+2\mathrm{y}-3}{\sqrt{5}}$( ∵ X is the distance of a point on hyperbola from 2x−y+4=0 and Y is the distance of a point on hyperbola from x+2y−3=0  )

So,required equation is $\frac{2{(2x-y+4)}^2}{5}-\frac{3{(x+2y-3)}^2}{5}=1$