Projectile - Minimum velocity to cross a point

Well we have to think this question as projectile on an incline. Join the points on from the origin to (a,b) and now you have to think like a projectile thrown with min vel. v can get this range. For vel to be min Range must be max which is √a²+b²in this case.
If (θ-angle of inclined plane, α-angle of projection with incline)

The formula for Range on an incline is =
2v²sinαcos(α+θ)/g cos²Î¸

For maximum it turns out to be
= v²/g(1+sinθ)
Since tanθ=ba, you can find out sinθ.

v=√g(√a²+b² +b)