Let $ f (x,y) = sin(ax + y^2)$ with $ a \in R$

Find the value of $a$ such that the tangent plane to the graph of $f$ in the point $(0,\sqrt\pi,0)$ goes through the point $(1,\sqrt\pi,5)$

Solution:

The tangent plane of $f$ exists so $f$ is differentiable which means that $f$ can be written as:

$ f (\overrightarrow{x}) = f(\overrightarrow{x\_0}) + <\overrightarrow{m},\overrightarrow{x}-\overrightarrow{x\_0}> + o(||\overrightarrow{x}-\overrightarrow{x\_0}||)$

where 

$y = f(\overrightarrow{x\_0}) + <\overrightarrow{m},\overrightarrow{x}-\overrightarrow{x\_0}>$ is the equation of the tangent plane of $f$ in the point $(\overrightarrow{x\_0},f(\overrightarrow{x\_0})$

and

$m = \nabla f(\overrightarrow{x\_0})$

So $\overrightarrow{x\_0} = (0,\sqrt\pi)$ :

$ f (\overrightarrow{x\_0}) = 0$

and 

$\nabla f(\overrightarrow{x}) = (a cos(ax+y^2),2ycos(ax+y^2),0)$

$m = \nabla f(\overrightarrow{x\_0}) = (-a,-2\sqrt\pi,0)$

Considering $\overrightarrow{x} = (x\_1, x\_2, x\_3)$ I have:

$y = f(\overrightarrow{x\_0}) + <\overrightarrow{m},\overrightarrow{x}-\overrightarrow{x\_0}>$

$5 = 0 + <(-a,-2\sqrt\pi,0),(x\_1,x\_2-\sqrt\pi,x\_3)>$

$5 = -ax\_1 - 2x\_2\sqrt\pi + 2\pi$

$5 = -a - 2\sqrt\pi\sqrt\pi + 2\pi$

$a = -5$ 

The value of $a$ is right, I would like to know if I solved it in the right way