Given Equation of Hyperbola as
$$\frac{(x-2y+3)^2}{20}-\frac{(2x+y-1)^2}{\frac{80}{9}}=1$$
we have axes equations as $x-2y+3=0$ and $2x+y-1=0$. But how to identify which is Transverse and which is conjugate?
Given Equation of Hyperbola as
$$\frac{(x-2y+3)^2}{20}-\frac{(2x+y-1)^2}{\frac{80}{9}}=1$$
we have axes equations as $x-2y+3=0$ and $2x+y-1=0$. But how to identify which is Transverse and which is conjugate?
Copyright © 2021 JogjaFile Inc.
Note that the hyperbola cuts only the transverse axis and not the conjugate axis. So only that axis which gives real points of intersection with hyperbola can be transverse axis.
So $2x+y-1 = 0$ gives real points when solved with hyperbola equation, it is the transverse axis!