Step 1: Know the touching rule for an ellipse.
A straight line $y = mx + c$ just touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at exactly one point when a special rule holds. That rule is $c^2 = a^2 m^2 + b^2$. So our whole job is to find $a^2$ and $b^2$.
Step 2: Why this rule works.
If a line meets an ellipse, we can put the line into the ellipse equation and get a quadratic in $x$. Touching means the line meets the curve at only one point, so that quadratic has equal roots. Setting its discriminant to zero gives exactly the rule above. We do not have to redo this each time.
Step 3: Put the ellipse in standard form.
The given ellipse is $9x^2 + 16y^2 = 144$. To compare it with the standard shape, divide every term by $144$.
\[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \]
Step 4: Simplify the fractions.
Simplify each part. $\frac{9}{144} = \frac{1}{16}$ and $\frac{16}{144} = \frac{1}{9}$. So the ellipse becomes $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Step 5: Read off $a^2$ and $b^2$.
Matching with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we see $a^2 = 16$ and $b^2 = 9$.
Step 6: Put these into the touching rule.
Substitute $a^2 = 16$ and $b^2 = 9$ into $c^2 = a^2 m^2 + b^2$.
\[ c^2 = 16m^2 + 9 \] So the line touches the ellipse when $c^2 = 16m^2 + 9$.
\[ \boxed{c^2 = 16m^2 + 9} \]