A line passing through $(21, 30)$ and normal to the curve $y = 2\sqrt{x}$. If $m$ is slope of the normal then $m + 6 = $
Step-by-Step Solution
Key Concept: The normal line equation at a point on the parabola combined with a condition that it passes through another point constrains the slope parameter.
The equation of the normal at $(a,a^2)$ is $y = mx - 2m - m^3$ (with $a = 1$). If the normal passes through $(21,30)$, then $30 = 21m - 2m - m^3$, giving $m^3 - 19m + 30 = 0$. Factoring: $(m+5)(m-2)(m-3) = 0$ yields $m = -5, 2, 3$. Since $(am^2, -2am)$ must lie on the parabola, only $m = -5$ works. Therefore $m + 6 = 1$.
Correct Answer: 1