Question

Let A, B and C be the vertices of a variable right angled triangle inscribed in the parabola $y^2 = 16x$. Let the vertex B containing the right angle be $(4, 8)$ and the locus of the centroid of $\Delta ABC$ be a conic $C_0$. Then three times the length of latus rectum of $C_0$ is _______.

Answer 1

Correct Answer: 16

Step 1: Understanding the Concept:
We parameterize the vertices of the right-angled triangle on the parabola. Using the slope condition for perpendicularity at vertex B, we can relate the parameters of vertices A and C. Then, substitute these relations into the centroid coordinates to find its locus $C_0$.

Step 2: Key Formula or Approach:
Parametric coordinates for $y^2 = 4ax$ are $(at^2, 2at)$. Here $a = 4$, so points are $(4t^2, 8t)$.
Slope of a chord joining $t_1, t_2$ is $\frac{2}{t_1 + t_2}$.
For perpendicular lines, $m_1 m_2 = -1$.
Centroid $G(h, k)$ has coordinates $h = \frac{\sum x_i}{3}, k = \frac{\sum y_i}{3}$.

Step 3: Detailed Explanation:
Given B $(4, 8)$, it corresponds to the parameter $t_1 = 1$.
Let A and C have parameters $t_2$ and $t_3$.
Since angle at B is $90^\circ$, the product of slopes of AB and BC is $-1$.
$m_{AB} = \frac{2}{t_1 + t_2} = \frac{2}{1 + t_2}$
$m_{BC} = \frac{2}{t_1 + t_3} = \frac{2}{1 + t_3}$
$\left(\frac{2}{1 + t_2}\right) \left(\frac{2}{1 + t_3}\right) = -1 \implies (1 + t_2)(1 + t_3) = -4$
Expanding: $1 + t_2 + t_3 + t_2 t_3 = -4 \implies t_2 t_3 = -5 - (t_2 + t_3)$.
Let the centroid be $G(h, k)$.
$h = \frac{4(1^2) + 4t_2^2 + 4t_3^2}{3} \implies t_2^2 + t_3^2 = \frac{3h}{4} - 1$.
$k = \frac{8(1) + 8t_2 + 8t_3}{3} \implies t_2 + t_3 = \frac{3k}{8} - 1$.
We know the algebraic identity: $(t_2 + t_3)^2 = t_2^2 + t_3^2 + 2t_2 t_3$.
Substitute the centroid relations and the perpendicularity constraint into this identity:
$\left( \frac{3k}{8} - 1 \right)^2 = \left( \frac{3h}{4} - 1 \right) + 2(-5 - (t_2 + t_3))$
$\frac{9k^2}{64} - \frac{3k}{4} + 1 = \frac{3h}{4} - 1 - 10 - 2\left(\frac{3k}{8} - 1\right)$
$\frac{9k^2}{64} - \frac{3k}{4} + 1 = \frac{3h}{4} - 11 - \frac{3k}{4} + 2$
$\frac{9k^2}{64} + 1 = \frac{3h}{4} - 9$
$\frac{9k^2}{64} = \frac{3h}{4} - 10 = \frac{3}{4} \left( h - \frac{40}{3} \right)$
Multiply by $\frac{64}{9}$:
$k^2 = \frac{64}{9} \cdot \frac{3}{4} \left( h - \frac{40}{3} \right) = \frac{16}{3} \left( h - \frac{40}{3} \right)$.
Replacing $(h, k)$ with $(x, y)$, the locus $C_0$ is the parabola:
$y^2 = \frac{16}{3} \left( x - \frac{40}{3} \right)$.
The length of the latus rectum of this new parabola is the coefficient of the linear $x$ term, which is $L.R. = \frac{16}{3}$.
We are asked for three times the length of the latus rectum:
$3 \times \frac{16}{3} = 16$.

Step 4: Final Answer:
The required value is 16.