1. **Problem statement:** Find the condition for which the line $y=mx+c$ touches the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1.$$ Then find the equation of the tangents with slope $m = -\frac{3}{5}$. 2. **Condition for tangency:** Substitute $y=mx+c$ into the ellipse equation: $$\frac{x^2}{25} + \frac{(mx+c)^2}{16} = 1.$$ Multiply both sides by 400 (LCM of 25 and 16) to clear denominators: $$16x^2 + 25(m^2x^2 + 2mcx + c^2) = 400.$$ Expand: $$16x^2 + 25m^2x^2 + 50mcx + 25c^2 = 400.$$ Group terms: $$ (16 + 25m^2)x^2 + 50mcx + (25c^2 - 400) = 0.$$ 3. **Tangency means the quadratic in $x$ has exactly one solution, so its discriminant $D=0$: $$D = (50mc)^2 - 4(16 + 25m^2)(25c^2 - 400) = 0.$$ Calculate discriminant: $$2500 m^2 c^2 - 4(16 + 25m^2)(25c^2 - 400) = 0.$$ Expand: $$2500 m^2 c^2 - 4(16 + 25m^2)25 c^2 + 4(16 + 25m^2)400 = 0.$$ $$2500 m^2 c^2 - 100 (16 + 25m^2) c^2 + 1600 (16 + 25m^2) = 0.$$ Group $c^2$ terms: $$c^2 (2500 m^2 - 100 (16 + 25m^2)) + 1600 (16 + 25m^2) = 0.$$ Simplify inside parentheses: $$2500 m^2 - 1600 - 2500 m^2 = -1600.$$ So: $$-1600 c^2 + 1600 (16 + 25m^2) = 0.$$ Divide both sides by 1600: $$-c^2 + 16 + 25 m^2 = 0.$$ Rearranged: $$c^2 = 16 + 25 m^2.$$ 4. **Equation of tangent lines with slope $m = -\frac{3}{5}$:** Calculate $c$: $$c^2 = 16 + 25 \left(-\frac{3}{5}\right)^2 = 16 + 25 \times \frac{9}{25} = 16 + 9 = 25.$$ So: $$c = \pm 5.$$ 5. **Final tangent equations:** $$y = -\frac{3}{5} x + 5$$ and $$y = -\frac{3}{5} x - 5.$$