The tangent to $x^2 = -12y$ is $kx = y + (-3)k^2$. Normal to $y^2 = 4x$ is $y = mx - 2m - m^3$.
Step-by-Step Solution
Key Concept: Finding common tangent and normal by equating slopes and solving the resulting equations gives the slope difference.
From the tangent to $x^2 = -12y$ we have $kx = y + (−3)k^2$. From the normal to $y^2 = 4x$ we have $y = mx − 2m − m^3$. Setting $k = m$ and $2m + m^3 = 3k^2$ gives $m(m^2 + 3 + 2) = 0 ⟹ m = −1, −2$ (m = 0 is rejected). The difference of slopes equals 1.
Correct Answer: 3