From a point $P$ three normals are drawn to the parabola $y^2 = 4ax$, such that the product of slopes of two of the normal is $p$. If the locus of $P$ is a part of the parabola, then $|p|$ equal to
Step-by-Step Solution
Key Concept: For three normals drawn from point P(h,k) to parabola y²=4ax with slopes m₁, m₂, m₃, use Vieta's formulas on the cubic equation k = mh - 2am - am³ to get m₁m₂m₃ = -k/a. If m₁m₂ = p, then m₃ = -k/(ap), and substituting back into the normal equation gives the locus. For this locus to be part of the original parabola y² = 4ax, coefficients must match, requiring |p| = 2.
For point $P(h,k)$, the normal slope equation is $k = mh - 2am - am^3$, giving roots $m_1, m_2, m_3$ satisfying $m_1m_2m_3 = -\frac{k}{a}$ and $m_1m_2 = p$. From these relations, $m_3 = -\frac{k}{ap}$. Substituting into the normal equation yields the locus $y^2 = ap^2x + (p-2)a^2p^2$. For this to equal $y^2 = 4ax$, we need $p^2 = 4$ and $p - 2 = 0$, giving $p = 2$.
Correct Answer: 2