Step 1: Put the hyperbola in standard form.
Divide $9x^2-16y^2=144$ by 144: \[ \frac{x^2}{16}-\frac{y^2}{9}=1\implies a^2=16,\ b^2=9 \]
Step 2: Find the latus rectum end in the third quadrant.
The ends of the latus rectum are $\left(\pm ae,\ \pm\frac{b^2}{a}\right)$, where $ae=\sqrt{a^2+b^2}=\sqrt{16+9}=5$. For the third quadrant both coordinates are negative: \[ (x_1,y_1)=\left(-5,\ -\tfrac{9}{4}\right) \]
Step 3: Recall the normal at a point.
The normal at $(x_1,y_1)$ to the hyperbola is \[ \frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2 \]
Step 4: Substitute the numbers.
\[ \frac{16x}{-5}+\frac{9y}{-\tfrac{9}{4}}=16+9\implies-\frac{16}{5}x-4y=25 \]
Step 5: Clear the fractions.
Multiply through by $-5$: \[ 16x+20y+125=0 \] So $a=16$, $b=20$, $c=125$.
Step 6: Compute the required ratio.
\[ \frac{b+c}{a}=\frac{20+125}{16}=\frac{145}{16} \] \[ \boxed{\tfrac{145}{16}} \]