Answer: \(a=3\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

At the point of contact, the line and curve have equal \(y\)-values:

\(ax^2-5x+2=(2a+1)x-10.\)

Bring all terms to one side:

\(ax^2-(2a+6)x+12=0.\)

For the line to be a tangent, this quadratic in \(x\) must have one repeated root.

So its discriminant is zero:

\(\big(-(2a+6)\big)^2-4(a)(12)=0.\)

Hence

\((2a+6)^2-48a=0.\)

Expanding gives

\(4a^2+24a+36-48a=0.\)

So

\(4a^2-24a+36=0.\)

Divide by \(4\):

\(a^2-6a+9=0.\)

Therefore

\((a-3)^2=0,\)

so

\(a=3.\)

This completes the solution and gives the required result.