Mathematics - Point Question with Solution | TestHub

$tx-2y-3t=0.......\left(1\right)$

$x-2ty+3=0......\left(2\right)$ ,

Multiply by $t$ and subtract from above equation, we get

$y\left(2t^2-2\right)=6t$

Now multiply by $t$ in first equation and subtract second, we get

$\left(t^2-1\right)x=3\left(t^2-1\right)$

$\Rightarrow x=3\frac{{(t}^2+1)}{t^2-1}$

$\Rightarrow \frac{x^2}{9}={\left(\frac{t^2+1}{t^2-1}\right)}^2 \& \frac{2y}{3}=\frac{2t}{t^2-1}$

$\Rightarrow \frac{4y^2}{9}=\frac{4t^2}{{\left(t^2-1\right)}^2}$

$\Rightarrow \frac{x^2}{9}-\frac{4y^2}{9}=1$

$\Rightarrow \frac{x^2}{9}-\frac{y^2}{\frac{9}{4}}=1$

$a^2=9;\Rightarrow a=3$

$b^2=\frac{9}{4}$$\Rightarrow b=\frac{3}{2}$
$\therefore$ Length of conjugate axis  $=2b=2\times \frac{3}{2}=3$

    Length of transverse axis$=2a= 6$

$e^2=1+\frac{\frac{9}{4}}{9}=1+\frac{1}{4}; e=\frac{\sqrt{5}}{2}$