The locus of the point of intersection of the straight lines, tx -...

tx - 2y - 3t = 0 . . . . . . . $\left(1\right)<br /><br />$x - 2ty + 3 = 0 . . . . . . \left(2\right)$,</p><p>Multiply by$t$and subtract from above equation, we get</p><p>$y\left(2{t}^{2} - 2\right) = 6t$</p><p>Now multiply by$t$in first equation and subtract second, we get</p><p>\left({t}^{2} - 1\right)x = 3 \left({t}^{2} - 1\right)$

$⇒ x = 3\dfrac{{( t}^{2} + 1 )}{{t}^{2} - 1}$

$⇒ \dfrac{{x}^{2}}{9} = {\left(\dfrac{{t}^{2} + 1}{{t}^{2} - 1}\right)}^{2} \text{\text{ and } } \dfrac{2y}{3} = \dfrac{2t}{{t}^{2} - 1}$

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

$⇒ \dfrac{{x}^{2}}{9} - \dfrac{4{y}^{2}}{9} = 1$

$⇒ \dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{\dfrac{9}{4}} = 1$

${a}^{2} = 9; ⇒ a=3$

${b}^{2} = \dfrac{9}{4}⇒ b = \dfrac{3}{2}$
$∴$ Length of conjugate axis $= 2b = 2 \times \dfrac{3}{2} = 3$

Length of transverse axis$= 2a = 6$

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