A point on the parabola $y^2=18x$ at which the ordinate increases at twice the rate of the abscissa is

Given the parabola equation $y^2=18x$.
Differentiating with respect to time $t$:
$2y\frac{dy}{dt}=18\frac{dx}{dt}\Rightarrow y\frac{dy}{dt}=9\frac{dx}{dt}$.
The problem states that the ordinate ($y$) increases at twice the rate of the abscissa ($x$):
$\frac{dy}{dt}=2\frac{dx}{dt}$.
Substituting this into the differentiated equation:
$y(2\frac{dx}{dt})=9\frac{dx}{dt}$
$2y=9\Rightarrow y=\frac{9}{2}$.
Now, find the $x$ -coordinate using the parabola equation:
$(\frac{9}{2}{)}^2=18x$
$\frac{81}{4}=18x\Rightarrow x=\frac{81}{4\times 18}=\frac{9}{4\times 2}=\frac{9}{8}$.
The point is $(\frac{9}{8},\frac{9}{2})$.