1. **State the problem:** Find the equation of the tangent line to the ellipse $$4x^2 + 9y^2 = 40$$ that has a slope of $$-\frac{2}{9}$$. 2. **Recall the ellipse equation and implicit differentiation:** The ellipse is given by $$4x^2 + 9y^2 = 40$$. To find the slope of the tangent line, differentiate implicitly with respect to $$x$$: $$\frac{d}{dx}(4x^2) + \frac{d}{dx}(9y^2) = \frac{d}{dx}(40)$$ which gives $$8x + 18y \frac{dy}{dx} = 0$$. 3. **Solve for $$\frac{dy}{dx}$$ (the slope of the tangent):** $$18y \frac{dy}{dx} = -8x \Rightarrow \frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}$$. 4. **Set the slope equal to the given slope:** $$-\frac{4x}{9y} = -\frac{2}{9}$$ Multiply both sides by $$9y$$: $$-4x = -2y$$ Simplify: $$4x = 2y \Rightarrow y = 2x$$. 5. **Substitute $$y = 2x$$ into the ellipse equation to find the points of tangency:** $$4x^2 + 9(2x)^2 = 40$$ $$4x^2 + 9 \times 4x^2 = 40$$ $$4x^2 + 36x^2 = 40$$ $$40x^2 = 40$$ $$x^2 = 1 \Rightarrow x = \pm 1$$. 6. **Find corresponding $$y$$ values:** For $$x=1$$, $$y=2(1)=2$$. For $$x=-1$$, $$y=2(-1)=-2$$. 7. **Write the tangent line equations using point-slope form:** Slope $$m = -\frac{2}{9}$$. For point $$(1,2)$$: $$y - 2 = -\frac{2}{9}(x - 1)$$ Simplify: $$y = -\frac{2}{9}x + \frac{2}{9} + 2 = -\frac{2}{9}x + \frac{20}{9}$$. For point $$( -1, -2)$$: $$y + 2 = -\frac{2}{9}(x + 1)$$ Simplify: $$y = -\frac{2}{9}x - \frac{2}{9} - 2 = -\frac{2}{9}x - \frac{20}{9}$$. **Final answer:** The equations of the tangent lines with slope $$-\frac{2}{9}$$ are $$y = -\frac{2}{9}x + \frac{20}{9}$$ and $$y = -\frac{2}{9}x - \frac{20}{9}$$.