Coordinates of the feet of normal drawn from the point $(7, 14)$ to the parabola $x^2 - 8x - 16y = 0$ is/are:
Step-by-Step Solution
Key Concept: Complete the square to identify the parabola's vertex, then use the normal equation parametrized by slope to find all feet of normals passing through a given point.
The parabola equation $x^2 - 8x - 16y = 0$ is rewritten as $(x-4)^2 = 16(y+1)$ by completing the square. Shifting origin to $(4, -1)$ gives $X^2 = 16Y$. The normal at point $(7, 14)$ has equation $X = mY - 8m - 4m^3$, which transforms to $(3, 1.5)$ in shifted coordinates. Solving the resulting cubic $(m-1)(2m-1)(2m+3) = 0$ yields $m = 1, \frac{1}{2}, -\frac{3}{2}$. Converting back to original coordinates gives feet of normals at $(-4, 3)$, $(0, 0)$, and $(16, 8)$.
Correct Answer: 1,2,4