JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 78

From the first line $t=\dfrac{2y}{x-3}$. Substituting into the second, $x+3=\dfrac{4y^2}{x-3}$, so
$$\frac{x^2}9-\frac{y^2}{9/4}=1,$$
a hyperbola with $b=\frac32$: the conjugate axis is 3. (Its eccentricity is $\frac{\sqrt5}2$, not $\sqrt5$.)