Two functions too trouble

Relevant wiki: Maximum value of a quadratic expression

Manipulating $g(y)$ ,

$g(y) = 9+7\cos ^{2} x + 24\sin x \cos x= 9(\sin ^{2}x + \cos ^{2} x) + 7\cos ^{2} x + 24\sin x \cos x =$

$(3\sin x)^{2} + (4\cos x)^{2} + 2(3\sin x)(4 \cos x) =$ $( 3sin x + 4cos x)^{2} \leq 25 \cdots \cdots (*)$

Now let's find maximum value of $f$ .

Substitute $t = x + \sqrt{x^{2} + 9} \implies x = \dfrac{t^{2}-9}{2t}$

$f(t) = -t^{2}+2at+9$

$\implies f_{max} = f(a) = a^{2}+9$

Maximum value of $f(t) = 25$ (by $*$ )

Therefore $a^{2}+9=25$

$\large \displaystyle \implies \boxed{a=4}$