$\text{ If }y=2x-3\text{ is a tangent to the parabola }{y}^2=4a\left(x-\frac{1}{3}\right)\text{ , then }a\text{ is equal to }$

$\text{ Solving }y=2x-3$

$\text{ and }{y}^2=4a\left(x-\frac{1}{3}\right)$

we have

$(2x-3{)}^2=4a\left(x-\frac{1}{3}\right)$

$\begin{array}{r}\text{ or }4x^2+9-12x=4ax-\frac{4a}{3}\\ \text{ or }4x^2-4(3+a)x+9+\frac{4a}{3}=0\end{array}$

This equation must have equal roots, i.e., D = 0

$\text{ or }16(3+a{)}^2-16\left(9+\frac{4a}{3}\right)=0$

$\begin{array}{l}\text{ or }9+a^2+6a=9+\frac{4a}{3}\\ \text{ or }{a}^2+\frac{14a}{3}=0\\ \text{ or }a=0\text{ or }a=-\frac{14}{3}\end{array}$