tx−2y−3t=0.......(1)
x−2ty+3=0......(2) ,
Multiply by t and subtract from above equation, we get
y(2t2−2)=6t
Now multiply by t in first equation and subtract second, we get
(t2−1)x=3(t2−1)
⇒x=3t2−1(t2+1)
\Rightarrow \frac{{x}^{2}}{9}={(\frac{{t}^{2}+1}{{t}^{2}-1})}^{2} & \frac{2y}{3}=\frac{2t}{{t}^{2}-1}
⇒94y2=(t2−1)24t2
⇒9x2−94y2=1
⇒9x2−49y2=1
a2=9;⇒a=3
b2=49⇒b=23
∴ Length of conjugate axis =2b=2×23=3
Length of transverse axis=2a=6
e2=1+949=1+41;e=25