1. **Problem statement:** We have a line with equation $$y = -5x + k + 5$$ and a curve with equation $$y = 7 - kx - x^2$$. The line is tangent to the curve, meaning they touch at exactly one point. 2. **Find the values of $k$ where the line is tangent to the curve:** Set the line and curve equal because at the point(s) of contact, $$y$$ is the same: $$-5x + k + 5 = 7 - kx - x^2$$ Rearranging: $$-5x + k + 5 - 7 + kx + x^2 = 0$$ $$x^2 + (k - 5)x + (k - 2) = 0$$ For tangency, this quadratic in $$x$$ has exactly one solution, so the discriminant is zero: $$\Delta = (k - 5)^2 - 4 \times 1 \times (k - 2) = 0$$ Simplify: $$(k - 5)^2 - 4(k - 2) = 0$$ $$k^2 - 10k + 25 - 4k + 8 = 0$$ $$k^2 - 14k + 33 = 0$$ Solve for $$k$$ using the quadratic formula: $$k = \frac{14 \pm \sqrt{14^2 - 4 \times 1 \times 33}}{2} = \frac{14 \pm \sqrt{196 - 132}}{2} = \frac{14 \pm \sqrt{64}}{2}$$ $$k = \frac{14 \pm 8}{2}$$ Two values: $$k = \frac{14 + 8}{2} = 11$$ and $$k = \frac{14 - 8}{2} = 3$$ 3. **For each value of $$k$$, find the equation of the tangent, the curve, and point of contact:** - For $$k=11$$: - Tangent line: $$y = -5x + 11 + 5 = -5x + 16$$ - Curve: $$y = 7 - 11x - x^2$$ - Find point of contact by substituting line into curve: $$-5x + 16 = 7 - 11x - x^2$$ Rearranged: $$x^2 + (11 - 5)x + (7 - 16) = 0$$ $$x^2 + 6x - 9 = 0$$ Since tangent, discriminant is zero: Confirm: $$\Delta = 6^2 - 4 \times 1 \times (-9) = 36 + 36 = 72\neq 0$$ Check: Since previous step found discriminant zero for the original quadratic, re-check the substitution carefully. Alternatively, use the point of tangency formula. Or use derivative approach: The derivative of the curve $$y = 7 - 11x - x^2$$ is $$y' = -11 - 2x$$. At contact point $$x=c$$, the slope equals the line's slope: $$-5 = -11 - 2c \Rightarrow -5 +11 = -2c \Rightarrow 6 = -2c \Rightarrow c = -3$$ Compute $$y$$: $$y = -5(-3) + 16 = 15 +16 = 31$$ So point of contact: $$(-3, 31)$$ - For $$k=3$$: - Tangent line: $$y = -5x + 3 + 5 = -5x + 8$$ - Curve: $$y = 7 - 3x - x^2$$ - Slope of curve: $$y' = -3 - 2x$$ Set equal to line slope $$-5$$: $$-5 = -3 - 2x \Rightarrow -5 + 3 = -2x \Rightarrow -2 = -2x \Rightarrow x=1$$ Find $$y$$: $$y = -5(1) + 8 = -5 + 8 = 3$$ Point of contact: $$(1,3)$$ **Final answers:** (i) $$k = 11$$ or $$k = 3$$ (ii) For $$k=11$$: - Tangent line: $$y = -5x + 16$$ - Curve: $$y = 7 - 11x - x^2$$ - Point of contact: $$(-3, 31)$$ For $$k=3$$: - Tangent line: $$y = -5x + 8$$ - Curve: $$y = 7 - 3x - x^2$$ - Point of contact: $$(1, 3)$$