Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
At any point $P$ on the parabola $y^2 - 2y - 4x + 5 = 0$, a tangent is drawn which meets the directrix at $Q$. If the locus of $R$ which divides $QP$ externally in the ratio $1:2$ is $(y-1)^2(x+1)+\lambda = 0$, then find $\lambda$.
Step-by-Step Solution
Key Concept: For a parabola with non-standard position, parametrize points and use directrix properties to find geometric relationships.
For the parabola $(y-1)^2 = 4(x-1)$, the tangent at $P(t^2+1, 2t+1)$ is $x - ty + t^2 - 1 = 0$. The foot of the directrix $x = 0$ gives point $Q(0, (t^2+t-1)/t)$. The foot of perpendicular $R$ from $P$ to directrix lies at $(-t^2 - 1, 2t+1)$. Using the constraint on $\lambda = 4$.
Correct Answer: 4