If the equation of parabola is $y^2 = 8x$, then locus of $P$ is:
Step-by-Step Solution
Key Concept: Three normals can be drawn from an external point to a parabola y² = 8x. Using the condition that the cubic equation in slope m has three real roots, and applying Vieta's formulas (m₁m₂ = -1), we eliminate the parameter to find the locus of point P from which normals are drawn.
For normals to $y^2 = 8x$ passing through point P, the cubic equation $2m^3 + (4-h)m = 0$ arises. By Vieta's formulas with roots $m_1, m_2, m_3$: $m_1m_2 = -1$ and $m_3 = \frac{k}{2}$. Substituting back into the root condition and solving yields $k^2 = 2(h-6)$, giving the locus $y^2 = 2(x-6)$.
Correct Answer: 2