1. **State the problem:** Find the equations of the tangent lines to the curve given by $$x^4y^2 + x^2y^4 = 26x^6 - 6$$ at $$x=1$$. 2. **Implicit differentiation:** Differentiate both sides with respect to $$x$$. Use the product rule and chain rule where necessary. Given: $$x^4y^2 + x^2y^4 = 26x^6 - 6$$ Differentiate term by term: $$\frac{d}{dx}(x^4y^2) + \frac{d}{dx}(x^2y^4) = \frac{d}{dx}(26x^6 - 6)$$ For $$x^4y^2$$: $$\frac{d}{dx}(x^4y^2) = 4x^3y^2 + x^4 \cdot 2y \frac{dy}{dx} = 4x^3y^2 + 2x^4y \frac{dy}{dx}$$ For $$x^2y^4$$: $$\frac{d}{dx}(x^2y^4) = 2xy^4 + x^2 \cdot 4y^3 \frac{dy}{dx} = 2xy^4 + 4x^2y^3 \frac{dy}{dx}$$ Right side: $$\frac{d}{dx}(26x^6 - 6) = 156x^5$$ 3. **Combine derivatives:** $$4x^3y^2 + 2x^4y \frac{dy}{dx} + 2xy^4 + 4x^2y^3 \frac{dy}{dx} = 156x^5$$ Group $$\frac{dy}{dx}$$ terms: $$2x^4y \frac{dy}{dx} + 4x^2y^3 \frac{dy}{dx} = 156x^5 - 4x^3y^2 - 2xy^4$$ Factor $$\frac{dy}{dx}$$: $$\frac{dy}{dx}(2x^4y + 4x^2y^3) = 156x^5 - 4x^3y^2 - 2xy^4$$ 4. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{156x^5 - 4x^3y^2 - 2xy^4}{2x^4y + 4x^2y^3}$$ 5. **Evaluate at $$x=1$$:** First find $$y$$ values at $$x=1$$ by substituting into original equation: $$1^4 y^2 + 1^2 y^4 = 26(1)^6 - 6$$ $$y^2 + y^4 = 20$$ Rewrite as: $$y^4 + y^2 - 20 = 0$$ Let $$z = y^2$$, then: $$z^2 + z - 20 = 0$$ Solve quadratic: $$z = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2}$$ Possible $$z$$ values: $$z = 4$$ or $$z = -5$$ (discard negative since $$y^2 \geq 0$$) So $$y^2 = 4 \Rightarrow y = \pm 2$$ 6. **Calculate slopes at points (1,2) and (1,-2):** For $$y=2$$: $$\frac{dy}{dx} = \frac{156(1)^5 - 4(1)^3(2)^2 - 2(1)(2)^4}{2(1)^4(2) + 4(1)^2(2)^3} = \frac{156 - 16 - 32}{4 + 32} = \frac{108}{36} = 3$$ For $$y=-2$$: $$\frac{dy}{dx} = \frac{156 - 4(1)(-2)^2 - 2(1)(-2)^4}{2(1)(-2) + 4(1)(-2)^3} = \frac{156 - 16 - 32}{-4 - 32} = \frac{108}{-36} = -3$$ 7. **Write tangent line equations:** Use point-slope form $$y - y_1 = m(x - x_1)$$ with $$x_1=1$$. For $$y=2$$ and slope $$3$$: $$y - 2 = 3(x - 1) \Rightarrow y = 3x - 3 + 2 = 3x - 1$$ For $$y=-2$$ and slope $$-3$$: $$y + 2 = -3(x - 1) \Rightarrow y = -3x + 3 - 2 = -3x + 1$$ **Final answer:** The tangent lines at $$x=1$$ are: $$y = 3x - 1$$ and $$y = -3x + 1$$