The equation of tangent to the parabola \(y^2\) = 4ax is y = mx + \(a\over m\).
Since it is drawn from point (2,3)
Therefore it lies on tangent y = mx + \(a\over m\).
\(\implies\) 3 = 2m + \(a\over m\)
\(\implies\) 3m = 2\(m^2\) + a
\(\implies\) 2\(m^2\) – 3m + a = 0
Now, Sum of slopes is \(3\over 2\). [ \(\because\) sum of roots = \(-b\over a\) ]