Question

Let c be the constant number such that c > 1. If the least area of the figure by the line passing through the point (1, c) with gradient 'm' and the parabola y = x² is 36 sq. units, find the value of (c² + m²).

Answer 1

104

Answer 2

The area $A$ between the parabola $y=x^2$ and the line $y=mx-m+c$ is given by: $A = \frac{1}{6} (m^2 - 4m + 4c)^{3/2}$ We are given that the least area is 36 sq. units. This implies: $A_{min} = 36$ $\frac{1}{6} (m^2 - 4m + 4c)_{min}^{3/2} = 36$ $(m^2 - 4m + 4c)_{min}^{3/2} = 216$ $(m^2 - 4m + 4c)_{min} = (216)^{2/3} = 36$ Let $E(m, c) = m^2 - 4m + 4c$. We need to find the minimum of $E(m, c)$ subject to $c > 1$. $E(m, c) = (m-2)^2 - 4 + 4c$ The minimum of $(m-2)^2$ is 0, occurring at $m=2$. Substituting $m=2$: $E(2, c) = (2-2)^2 - 4 + 4c = 4c - 4$ Since the minimum value of $E$ is 36, we set $E(2, c) = 36$: $4c - 4 = 36$ $4c = 40$ $c = 10$ The values that yield the minimum area are $m=2$ and $c=10$. These satisfy $c>1$. We need to find $c^2 + m^2$: $c^2 + m^2 = 10^2 + 2^2 = 100 + 4 = 104$