If one zero of the polynomial

$(a+9) x^{2}-13 x+6 a=0$

Here, $A=\left(a^{2}+9\right), B=13$ and $C=6 a$

Let $\alpha$ and $\frac{1}{\alpha}$ be the two zeroes.

Then, product of the zeroes $=\frac{C}{A}$

$=>\alpha \cdot \frac{1}{\alpha}=\frac{6 a}{a^{2}+9}$

$=>1=\frac{6 a}{a^{2}+9}$

$=>a^{2}+9=6 a$

$=>a^{2}-6 a+9=0$

$=>a^{2}-2 \times a \times 3+3^{2}=0$

$=>(a-3)^{2}=0$

$=>a-3=0$

$=>a=3$