The point $P(9/2, 6)$ ;lies on the parabola $y^2=4ax$, then parameter of the point P -

As point $P\left( \dfrac { 9 }{ 2 } ,6 \right)$ lies on the parabola ${ y }^{ 2 }=4ax$, it must satisfy the equation of parabola.

$\therefore { 6 }^{ 2 }=4a\left( \dfrac { 9 }{ 2 } \right)$

$\therefore 36=18a$

$\therefore a=2$

Now, point P which lies on parabola has coordinates $P\left( a{ t }^{ 2 },2at \right)$

Where, t is a parameter

$\therefore a{ t }^{ 2 }=\dfrac { 9 }{ 2 }$

$\therefore 2{ t }^{ 2 }=\dfrac { 9 }{ 2 }$

$\therefore { t }^{ 2 }=\dfrac { 9 }{ 4 }$

$\therefore { t }=\dfrac { 3 }{ 2 }$