1. **Problem statement:** We have a line with equation $y = mx + c$ and a curve with equation $xy = 16$. (a) The line is tangent to the curve. We need to express $m$ in terms of $c$. (b) Given $m = -4$, find the values of $c$ for which the line intersects the curve at two distinct points. --- 2. **Expressing $m$ in terms of $c$ when the line is tangent to the curve:** - Substitute $y = mx + c$ into $xy = 16$: $$x(mx + c) = 16$$ which simplifies to $$m x^2 + c x - 16 = 0$$ - This is a quadratic equation in $x$. For the line to be tangent to the curve, this quadratic must have exactly one solution. - The discriminant $\Delta$ of the quadratic $a x^2 + b x + c = 0$ is $\Delta = b^2 - 4ac$. - Here, $a = m$, $b = c$, and $c = -16$ (constant term). - Set the discriminant to zero for tangency: $$\Delta = c^2 - 4 \times m \times (-16) = 0$$ $$c^2 + 64 m = 0$$ - Solve for $m$: $$m = -\frac{c^2}{64}$$ --- 3. **Finding $c$ values for two distinct intersections when $m = -4$:** - Substitute $m = -4$ into the quadratic: $$-4 x^2 + c x - 16 = 0$$ - The discriminant is: $$\Delta = c^2 - 4 \times (-4) \times (-16) = c^2 - 256$$ - For two distinct points, $\Delta > 0$: $$c^2 - 256 > 0$$ $$c^2 > 256$$ - Taking square roots: $$|c| > 16$$ - So the set of $c$ values is: $$c < -16 \quad \text{or} \quad c > 16$$ --- **Final answers:** (a) $$m = -\frac{c^2}{64}$$ (b) $$c < -16 \quad \text{or} \quad c > 16$$