If three normals are drawn from the point $(c, 0)$ to the parabola $y^2 = 4x$ and two of which are perpendicular, then the value of $c$ is equal to
Step-by-Step Solution
Key Concept: The normal to a parabola passes through a specific point if the slope equation is satisfied; equating the normal equation evaluated at that point yields constraints on the parabola parameter.
The equation of the normal in slope form is $y = mx - 2m - m^3$, which passes through $(-c, 0)$. Substituting: $0 = m(-c) - 2m - m^3$, giving $0 = mc - (2 - c)m = m^3 + (2 - c)m = 0$. Either $m = 0$ or $m^2 + (2 - c) = 0$. For the latter, $m = \pm\sqrt{c - 2}$. From $m_1 m_2 = \frac{4c}{-1} = -1$ and solving, we get $c = 3$.
Correct Answer: 1